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QUESTION 1 Reaction of pentanoic acid with thionyl chloride produces compound B followed by addition of two equivalents of am

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Answer #1

Solution :

1) Option C (Pentamide)

CH3CH2CH2CH2COOH + SOCl2 ==> CH3CH2CH2CH2COCl

CH3CH2CH2CH2COCl + 2NH3 ===>  CH3CH2CH2CH2CONH2 + NH4Cl

Thus, final product is pentamide.

2) Option A (hexylamine)

Pentanoic acid + LiAlH4 + H3O+ === Pentanol

Pentanol + PBr3 = Pentyl bromide

Pentyl bromide + NaCN =>>> CH3CH2CH2CH2CH2CN

CH3CH2CH2CH2CH2CN + LiAlH4 + H2O ==> CH3CH2CH2CH2CH2CH2NH2 (hexylamine)

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