organic compound has molecular formula CH11(A). A on heating with soda lime yields a hydrocarbon BCH) Bonblorination yields a single isomeric alkyl chloride C(CH.Cl). Deduce the structures of A, B and C
**Solution.** We will proceed from two cases, because it is not entirely clear what is meant by the expression "the only isomeric alkyl chloride".
1) We understand this expression in such a way that one alkyl chloride is formed during chlorination. Then B is methane (CH4). When chlorination is the reaction: CH4+Cl2=CH3Cl+HCl. Then substance C is chloromethane (CH3Cl). Since substance A initially reacts with soda lime, then A is acetic acid (CH3COOH), and reaction: CH3COOH+NaOH=NaHCO3+CH4.
2) We understand this expression as the formation of two isomeric chlorinated alkanes, that is, one chlorinated alkane has one single isomer. Then B is 2-methylpropane. Then the chlorination reaction will go in two ways:
H3C−CH(CH3)−CH3+Cl2=H3C−C(Cl)(CH3)−CH3 (2-chloro-2-methylpropane)+HCl;H3C−CH(CH3)−CH3+Cl2=H3C−CH(CH3)−CH2Cl (1-chloro-2-methylpropane)+HCl. Then substance C can be either chloralkane from the first reaction or chloralkan from the second reaction.
Substance A will then have the formula: H3C−CH(CH3)−CH2−COOH (3-methylbutanoic acid), and reaction with soda lime: H3C−CH(CH3)−CH2−COOH+NaOH=NaHCO3+H3C−CH(CH3)−CH3.
**Answer:** 1) A – acetic acid (CH3COOH), B – methane (CH4), C – chloromethane (CH3Cl);
2) A – 3-methylbutanoic acid (H3C−CH(CH3)−CH2−COOH), B – 2-methylpropane (H3C−CH(CH3)−CH3), C – H3C−C(Cl)(CH3)−CH3 (2-chloro-2-methylpropane) or H3C−CH(CH3)−CH2Cl (1-chloro-2-methylpropane).
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#339914 on Dec 2023