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JEE Advanced levelSingle correct optionDifficulty 3 of 3nc051

$H X_{2} NCH X_{2} CH X_{2} NHCH X_{3}$ is treated with excess $(CH X_{3} CO) X_{2} O$ (one acetyl group enters on each nitrogen bearing N-H, and the amides formed are not acylated further). The product is then reduced with excess $LiAlH X_{4}$ . The final compound is:

A $CH X_{3} CONHCH X_{2} CH X_{2} N (CH X_{3}) COCH X_{3}$
B $CH X_{3} CH X_{2} NHCH X_{2} CH X_{2} NHCH X_{3}$
C $CH X_{3} CH X_{2} NHCH X_{2} CH X_{2} N (CH X_{3}) CH X_{2} CH X_{3}$
D $CH X_{3} CH X_{2} N (CH X_{3}) CH X_{2} CH X_{2} N (CH X_{3}) CH X_{2} CH X_{3}$

Answer: C

The primary

NH X_{2}

becomes

NHCOCH X_{3}

and the secondary

NHCH X_{3}

becomes

N (CH X_{3}) COCH X_{3}

, giving a diamide.

LiAlH X_{4}

reduces each amide carbonyl to

CH X_{2}

, so each

COCH X_{3}

becomes an ethyl group on the same nitrogen. The product is

CH X_{3} CH X_{2} NHCH X_{2} CH X_{2} N (CH X_{3}) CH X_{2} CH X_{3}

; no extra methyl group is added.

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