CH_3-C(=O)-OH xrightarrow{NaH^{14}CO_3} (A) ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Both acetic acid $(CH_3COOH)$ and methanesulfonic acid $(CH_3SO_3H)$ are stronger acids than carbonic acid $(H_2CO_3)$. Therefore,they react with

Vedclass · Accepted Answer

$Ph-^{14}COOH$ and $Ph-^{14}COOH$

Vedclass · Answer

(C) Both acetic acid $(CH_3COOH)$ and methanesulfonic acid $(CH_3SO_3H)$ are stronger acids than carbonic acid $(H_2CO_3)$. Therefore,they react with $NaH^{14}CO_3$ to release $^{14}CO_2$ gas. Reaction $1$: $CH_3COOH + NaH^{14}CO_3 \rightarrow CH_3COONa + H_2O + ^{14}CO_2 \uparrow (A)$ Reaction $2$: $CH_3SO_3H + NaH^{14}CO_3 \rightarrow CH_3SO_3Na + H_2O + ^{14}CO_2 \uparrow (C)$ Both $(A)$ and $(C)$ are $^{14}CO_2$. When $^{14}CO_2$ reacts with Grignard reagent $(PhMgBr)$ followed by hydrolysis: $PhMgBr + ^{14}CO_2$ $\rightarrow Ph-^{14}COOMgBr$ $\xrightarrow{H_3O^{+}} Ph-^{14}COOH$ Thus,both $(B)$ and $(D)$ are $Ph-^{14}COOH$.

List-$I$ (Chemical Reaction)	List-$II$ (Reagent used)
$(a)$ $CH_{3}COOCH_{2}CH_{3} \rightarrow CH_{3}CH_{2}OH$	$(i)$ $CH_{3}MgBr / H_{3}O^{+}$ ($1$ equivalent)
$(b)$ $CH_{3}COOCH_{3} \rightarrow CH_{3}CHO$	$(ii)$ $H_{2}SO_{4} / H_{2}O$
$(c)$ $CH_{3}C \equiv N \rightarrow CH_{3}CHO$	$(iii)$ $DIBAL-H / H_{2}O$
$(d)$ $CH_{3}C \equiv N \rightarrow CH_{3}COCH_{3}$	$(iv)$ $SnCl_{2}, HCl / H_{2}O$

$CH_3-C(=O)-OH$ $\xrightarrow{NaH^{14}CO_3} (A) (\text{gas})$ $\xrightarrow[(ii) H_3O^{+}]{(i) PhMgBr} (B)$
$CH_3-S(=O)_2-OH$ $\xrightarrow{NaH^{14}CO_3} (C) (\text{gas})$ $\xrightarrow[(ii) H_3O^{+}]{(i) PhMgBr} (D)$
Product $(B)$ and $(D)$ in the above reaction are:

Similar Questions

The product $(C)$ of the given reaction sequence is:

The major product of the following reaction sequence is

The correct acidity order of the following compounds is:

Give the conversion of the following:
$(i)$ Phthalimide from $o-xylene$
$(ii)$ Ethanol from Ethanoic anhydride

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$CH_3-C(=O)-OH$ $\xrightarrow{NaH^{14}CO_3} (A) (\text{gas})$ $\xrightarrow[(ii) H_3O^{+}]{(i) PhMgBr} (B)$ $CH_3-S(=O)_2-OH$ $\xrightarrow{NaH^{14}CO_3} (C) (\text{gas})$ $\xrightarrow[(ii) H_3O^{+}]{(i) PhMgBr} (D)$ Product $(B)$ and $(D)$ in the above reaction are: