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class="mark-page-favorite pull-right" data-pid="910" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../342/organic-chemistry.html">Organic Chemistry</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Functional group chemistry</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 60 minutes"><i class="fa fa-clock-o"></i> 60'</span> </ol> <article id="main-article"> <p><img alt="" src="../../images/organic-chemistry/funcgrps-1.jpg" style="width: 160px; height: 116px; float: left;">Concept wise, this is not the most challenging section of the course, but there is so much detail that it can seem very daunting. The more you commit to memory the more confident you will feel with the organic section of the course. An organic compound is classified according to the specific functional groups that are present. Note that a distinction is made between class (type of compound - e.g. alcohol) and functional group (reactive site in compound - e.g. hydroxyl). <strong>Use the flash cards to</strong> <strong>learn the vocabulary and conventions used in organic chemistry thoroughly</strong>. Each class of organic compounds is associated with characteristic reactions and you should <strong>use the revision cards and practice questions to drill yourself in recognising these</strong>. Pay particular attention to what is happening to the formulae and functional groups as reactions take place.</p> <hr class="hidden-separator"> <div class="panel panel-has-colored-body panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <p>Ensure you are confident using the terms below and learn the asterisked* definitions</p> <p><strong>Hydrocarbon*</strong>, Saturated, Unsaturated, <strong>Free radical*</strong>, <strong>Nucleophile*</strong>, <strong>Electrophile*</strong>, <strong>Homolytic bond fission*</strong>, <strong>Heterolytic bond fission*</strong></p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="1" data-mode="" data-topics="631" data-subject-id="7" data-n-flashcards="8" style="text-align:center">Show flashcards</a></div><hr> </div> </div> </div> </div> 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data-index="12"><img title="Click to view" src="../../../std-galleries/7-220/screenshot-2020-08-23-at-220931-thumb128.jpg"><li><li data-index="13"><img title="Click to view" src="../../../std-galleries/7-220/screenshot-2020-08-23-at-220949-thumb128.jpg"><li></li></ol> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test yourself</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <div class="tib-quiz" data-stats="7-547-910"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following is a saturated hydrocarbon?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>OH</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>CH(CH<sub>3</sub>)CH<sub>3</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>COCH<sub>3</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>CHCH<sub>2</sub></span></label> </p></div><div class="q-explanation"><p>A hydrocarbon is a compound of carbon and hydrogen <strong>only</strong>. Saturated compounds contain only single carbon-carbon bonds (no double bonds).</p><p>CH<sub>3</sub>CH<sub>2</sub>CHCH<sub>2 </sub>is but-1-ene and is unsaturated (but is a hydrocarbon). CH<sub>3</sub>COCH<sub>3 </sub>and CH<sub>3</sub>CH<sub>2</sub>OH both contain oxygen and therefore are not hydrocarbons.</p><p><strong>CH<sub>3</sub>CH<sub>2</sub>CH(CH<sub>3</sub>)CH<sub>3 </sub></strong>is methylbutane, a saturated hydrocarbon and is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following statements explain why alkanes are relatively unreactive?</p><p><strong>1: </strong>They have strong C−H and C−C bonds</p><p><strong>2: </strong>The bonds are non-polar</p><p><strong>3: </strong>They have delocalised electrons</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 3 only</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1 and 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p></div><div class="q-explanation"><p>Alkanes are relatively unreactive because they have strong carbon-carbon and carbon-hydrogen bonds that require a lot of energy to break. In addition, the bonds are all non-polar, which means they are not vulnerable to attack by most chemical reagents.</p><p>Alkanes do not have delocalised electrons.</p><p>Therefore <strong>1 and 2 only </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the sum of all coefficients when the following equation (the complete combustion of pentane) is balanced using the smallest possible whole numbers?</p><p>___C<sub>5</sub>H<sub>12 </sub>+ ___O<sub>2 </sub>→ ___CO<sub>2</sub> + ___H<sub>2</sub>O</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 33<span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>20</span></label> </p><p><label class="radio"> <input type="radio"> <span>19</span></label> </p><p><label class="radio"> <input type="radio"> <span>35</span></label> </p></div><div class="q-explanation"><p>C<sub>5</sub>H<sub>12 </sub>+ 8O<sub>2 </sub>→ 5CO<sub>2</sub> + 6H<sub>2</sub>O is the balanced equation, so the total sum is 20 - don't forget that C<sub>5</sub>H<sub>12</sub> does have a '1' in front of it, although by convention we don't actually write that into the equation.</p><p>(When balancing combustion reactions of hydrocarbons, first balance out carbon and hydrogen atoms by adding coefficients (numbers) in front of CO<sub>2 </sub>and H<sub>2</sub>O, after that balance the oxygen atoms and the hydrocarbon as required.)</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the product of the reaction between propene (CH<sub>3</sub>CHCH<sub>2</sub>) and bromine?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>Br</span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CBrCHBr</span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>CHBr<sub>2</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>CH<sub>3</sub>CHBrCH<sub>2</sub>Br</span></label> </p></div><div class="q-explanation"><p>Propene contains a double carbon-carbon bond (C=C) that will readily undergo an addition reaction with bromine. The bromine molecule adds across the double bond, with one atom attaching to each carbon atom of the double bond.</p><p>Thus <strong>CH<sub>3</sub>CHBrCH<sub>2</sub>Br </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the major organic product at the end of the reaction when ethanol (CH<sub>3</sub>CH<sub>2</sub>OH) is refluxed with acidified potassium dichromate(VI) (H<sup>+</sup>/K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub>)?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>CO<sub>2</sub>H</span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CHO</span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>COOCH<sub>3</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>CH<sub>3</sub>CO<sub>2</sub>H</span></label> </p></div><div class="q-explanation"><p>Ethanol is a primary alcohol. When oxidised primary alcohols produce an aldehyde initially (must be distilled off immediately as formed) that then readily reacts further to form a carboxylic acid (under reflux).</p><p>Ethanol will therefore form ethanoic acid (CH<sub>3</sub>CO<sub>2</sub>H).</p><p>CH<sub>3</sub>CHO is an aldehyde (ethanal), CH<sub>3</sub>CH<sub>2</sub>CO<sub>2</sub>H is propanoic acid (too many carbons), CH<sub>3</sub>COOCH<sub>3 </sub>is methyl ethanoate, an ester.</p><p><strong>CH<sub>3</sub>CO<sub>2</sub>H</strong> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following could be formed during the incomplete combustion of octane fuel (C<sub>8</sub>H<sub>18</sub>)?</p><p><strong>1: </strong>Carbon monoxide</p><p><strong>2: </strong>Hydrogen</p><p><strong>3: </strong>Carbon</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p></div><div class="q-explanation"><p>Alkanes will completely combust in excess oxygen, producing carbon dioxide and water. If there is insufficient oxygen then carbon monoxide and carbon (soot) may be formed. Hydrogen gas will never be formed (it is too reactive especially at high temperature).</p><p>Therefore <strong>1 and 3 only </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following statements about the free radical substitution reaction of methane and chlorine are true?</p><p><strong>1: </strong>Chlorine undergoes homolytic fission</p><p><strong>2: </strong>Ultraviolet light is required to produce the free radicals</p><p><strong>3: </strong>Two free radicals are produced in the termination step</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>1 and 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p></div><div class="q-explanation"><p>Typically in free radical substitution an alkane reacts with a halogen and hydrogen atoms are readily substituted with halogen atoms.</p><p><strong>Initiation</strong> involves UV light breaking the halogen bond (homolytic fission) and producing two halogen free radicals (species with an unpaired electron - an <strong>odd</strong> number of electrons).</p><p><strong>Propogation</strong> involves radicals (odd) reacting with non-radical species (even) to produce a radical (odd) and a non-radical (even).</p><p><strong>Termination</strong> involves two free radicals (odd) coming together to produce a non-radical (even).</p><p>Therefore <strong>1 and 2 only </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>Ethene can react to form 1-chloroethane. Which reagent is best used to react with ethene to produce this product?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>UV light</span></label> </p><p><label class="radio"> <input type="radio"> <span>NaCl</span></label> </p><p><label class="radio"> <input type="radio"> <span>Cl<sub>2</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>HCl</span></label> </p></div><div class="q-explanation"><p>Ethene (CH<sub>2</sub>CH<sub>2)</sub> is an alkene with a double C=C bond that readily reacts through addition reactions.</p><p>1-chloroethane has formula CH<sub>3</sub>CH<sub>2</sub>Cl. HCl has been added to ethene, therefore <strong>HCl</strong> is the correct answer.</p><p>Addition of chlorine would produce 1,2-dichloroethane (CH<sub>2</sub>ClCH<sub>2</sub>Cl). Sodium chloride will not react with ethene (under any normal conditions) and exposure to UV light without a source of chlorine will not lead to a suitable product - even with chlorine this reaction would not be best used as the products would be very variable.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which monomer would be used to produce <em>poly(chloroethene)</em> using addition polymerisation?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>CHClCH<sub>2</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CHClCHCl</span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>2</sub>ClCH<sub>3</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>2</sub>ClCH<sub>2</sub>Cl</span></label> </p></div><div class="q-explanation"><p>To produce <em>poly(chloroethene) </em>the monomer needs to be <strong>chloroethene</strong>. The formula of chloroethene is <strong>CHClCH<sub>2</sub></strong>, which is therefore the correct answer. This molecule has a double C=C bond and one chlorine atom.</p><p>The other options either have no double C=C bond (CH<sub>2</sub>ClCH<sub>3</sub>) or<em> two</em> chlorine atoms (CHClCHCl) or both (CH<sub>2</sub>ClCH<sub>2</sub>Cl).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>What colour change is seen when a tertiary alcohol is oxidised (under reflux) with acidified potassium dichromate(VI) (H<sup>+</sup>/K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub>)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Purple to colourless</span></label> </p><p><label class="radio"> <input type="radio"> <span>Orange to green</span></label> </p><p><label class="radio"> <input type="radio"> <span>Pink to colourless</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>No change</span></label> </p></div><div class="q-explanation"><p>Tertiary alcohols cannot be oxidised with acidified potassium dichromate(VI) so no reaction will take place, and there will be no colour change (the reaction mixture will remain orange).</p><p>Primary and secondary alcohols can be oxidised (to aldehydes or carboxylic acids and to ketones respectively) and teh colour change would be orange to green.</p><p>Therefore <strong>No change</strong> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">11</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the correct formula for the organic product of the reaction between propanoic acid and ethanol in the presence of an acid catalyst?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>CO<sub>2</sub>CH<sub>2</sub>CH<sub>3</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>CO<sub>2</sub>CH<sub>3</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CO<sub>2</sub>CH<sub>2</sub>CH<sub>3</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CO<sub>2</sub>CH<sub>2</sub>CH<sub>2</sub>CH<sub>3</sub></span></label> </p></div><div class="q-explanation"><p>This is an esterification reaction. Esters are named by the 'alcohol bit' first then the 'acid bit', that is firstly the carbon chain section that is directly attached to the oxygen atom by a single bond, then the carbon chain section that includes the C=O.</p><p><img alt="" src="../../images/organic-chemistry/ester-naming.png" style="width: 240px; height: 141px;"></p><p>Propanoic acid and ethanol will form the ester <strong>ethyl propanoate</strong>, the formula of which is <strong>CH<sub>3</sub>CH<sub>2</sub>CO<sub>2</sub>CH<sub>2</sub>CH<sub>3</sub></strong>. The other compounds are all esters, but propyl ethanoate (CH<sub>3</sub>CO<sub>2</sub>CH<sub>2</sub>CH<sub>2</sub>CH<sub>3</sub>), methyl propanoate (CH<sub>3</sub>CH<sub>2</sub>CO<sub>2</sub>CH<sub>3</sub>) and ethyl ethanoate (CH<sub>3</sub>CO<sub>2</sub>CH<sub>2</sub>CH<sub>3</sub>).</p><p>Therefore the answer is <strong>CH<sub>3</sub>CH<sub>2</sub>CO<sub>2</sub>CH<sub>2</sub>CH<sub>3</sub></strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">12</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following are nucleophiles?</p><p><strong>1: </strong>OH<sup>−</sup></p><p><strong>2: </strong>NH<sub>3</sub></p><p><strong>3: </strong>CH<sub>4</sub></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 only</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1 and 2 only</span></label> </p></div><div class="q-explanation"><p>Nucleophiles are electron pair donors. In order for a species to be a nucleophile it must have a lone (non-bonding) pair of electrons. The hydroxide ion (OH<sup>−</sup>) has three lone pairs on the oxygen atom, ammonia (NH<sub>3</sub>) has one lone pair on the nitrogen atom. Both OH<sup>−</sup> and NH<sub>3</sub> can therefore behave as nucleophiles.</p><p>Methane (CH<sub>4</sub>) has no lone (non-bonding) pairs and therefore cannot behave as a nucelophile.</p><p>Therefore <strong>1 and 2 only </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">13</div><div class="exercise shadow-bottom"><div class="q-question"><p>Chloroethane (CH<sub>3</sub>CH<sub>2</sub>Cl) reacts to produce ethanol (CH<sub>3</sub>CH<sub>2</sub>OH). What reagent is needed for this reaction and what is the type of reaction taking place?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>NaOH<sub>(aq)</sub> / nucleophilic addition</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>NaOH<sub>(aq)</sub> / nucleophilic substitution</span></label> </p><p><label class="radio"> <input type="radio"> <span>Conc. H<sub>2</sub>PO<sub>4(aq)</sub> / nucleophilic substitution</span></label> </p><p><label class="radio"> <input type="radio"> <span>Conc. H<sub>2</sub>SO<sub>4(aq)</sub> / nucleophilic addition </span></label> </p></div><div class="q-explanation"><p>Halogenoalkanes will undergo nucleophilic substitution reactions will sodium hydroxide in water, to produce the corresponding alcohol. Thus chloroethane reacts with NaOH<sub>(aq)</sub> in a nucleophilic substitution reaction to produce ethanol.</p><p>Halogenoalkanes have no double bonds and cannot undergo addition reactions. The reagent conc. sulfuric acid (H<sub>2</sub>SO<sub>4(aq)</sub>) is a common reagent (and catalyst) but will not react to produce the alcohol in this reaction.</p><p>Therefore the correct answer is<strong> NaOH<sub>(aq)</sub> / nucleophilic substitution</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">14</div><div class="exercise shadow-bottom"><div class="q-question"><p>Benzene can react through an electrophilic substitution reaction to produce nitrobenzene.</p><p>Which of the following statements are correct with respect to this reaction?</p><p><strong>1: </strong>The electrophile is NO<sub>2</sub><sup>+</sup></p><p><strong>2: </strong>A hydrogen atom on the benzene ring is lost in the process</p><p><strong>3: </strong>Nitrobenzene has the formula C<sub>6</sub>H<sub>5</sub>NO<sub>2</sub></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>1, 2 and 3</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 3 only</span></label> </p></div><div class="q-explanation"><p>An electrophile is an electron pair acceptor. Electrophiles are electron deficient (often with a positive charge). Benzene will react, via an electrophilic substitution reaction with an electrophile which will replace a hydrogen atom on the benzene ring.</p><p>The nitration of benzene reaction: C<sub>6</sub>H<sub>6</sub> + NO<sub>2</sub><sup>+</sup> → C<sub>6</sub>H<sub>5</sub>NO<sub>2</sub> + H<sup>+</sup></p><p>Therefore <strong>1, 2 and 3 </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> 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