Consider the following reaction. The major pr | vedclass

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(A) Step $(i)$: $NaBH_{4}$ in $MeOH$ selectively reduces the aldehyde group $(-CHO)$ to a primary alcohol $(-CH_{2}OH)$. The cyclic amide (lactam) rem

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(A) Step $(i)$: $NaBH_{4}$ in $MeOH$ selectively reduces the aldehyde group $(-CHO)$ to a primary alcohol $(-CH_{2}OH)$. The cyclic amide (lactam) remains unaffected.Step (ii): $NaOH(aq.), \Delta$ causes the alkaline hydrolysis of the cyclic amide (lactam). The ring opens to form the corresponding amino acid salt (carboxylate).Step (iii): $H_{3}O^{+}$ protonates the carboxylate to form a carboxylic acid $(-COOH)$ and the amine to form an ammonium salt $(-NH_{2}^{+}-)$.The final product is a linear amino acid derivative with a primary alcohol group: $HOOC-(CH_{2})_{4}-NH-CH(CH_{3})CH_{2}OH$.

Consider the following reaction. The major product $(P)$ formed is:

Similar Questions

Which of the following reactions is called the 'Bouveault-Blanc reduction'?

The descending order of acidity for the following carboxylic acids is:
$A.$ $CH_3COOH$
$B.$ $F_3C-COOH$
$C.$ $ClCH_2-COOH$
$D.$ $FCH_2-COOH$
$E.$ $BrCH_2-COOH$
Choose the correct answer from the options given below:

Arrange the following in the increasing order of their acidic strength: $CH_{3}COOH, CH_{2}ClCOOH, CHCl_{2}COOH, CCl_{3}COOH$.

Statement-$I$: Most carboxylic acids exist as dimers in vapour phase or in aprotic solvents due to intermolecular hydrogen bonding.
Statement-$II$: Higher carboxylic acids are practically insoluble in water due to the increased hydrophobic interaction of the hydrocarbon part.

Which one of the following orders of acid strength is correct?

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