Question

$CH \equiv CH\xrightarrow[{HgS{O_4}(1\% )}]{{{H_2}S{O_4}(40\% )}}(A)\xrightarrow{{{K_2}C{r_2}{O_7}/{H^ + }}}(B)\xrightarrow{{(Ca{{(oH)}_2}}}(C)\xrightarrow{\vartriangle }(D)$
In this sequence of reactions, compound (D) is
A) Acetic Anhydride
B) Acetaldehyde
C) Acetone
D) Ethanol

Answer 1

All reactions are the preparation or the chemical properties of that substituent compound. To start with it is the preparation of ketone. Also, it will be the chemical property of that compound. So, we can solve the sequence of reactions keeping that in mind.

Complete step-by-step answer:
We can see that hydration of alkynes in presence of 40% sulphuric acid and of mercuric sulphate yield ketones except for acetylene which yields aldehyde. Though this is used to prepare ketone, we get aldehyde since we are using acetylene.

$CH \equiv {\text{CH}}\xrightarrow[{Hg{{(SO)}_4}\left( {1\% } \right)}]{{{H_2}{{(SO)}_4}\left( {40\% } \right)}}{\text{C}}{{\text{H}}_2} = {\text{CH}} - {\text{OH}}\xrightarrow{{{\text{Tautomerism}}}}{\text{C}}{{\text{H}}_3} - {\text{CHO}}$

We can see that the acetylene when reacted with $40\%$ sulphuric acid and $1\%$ of mercuric sulphate first gives us $prop - 1 - en - 2 - ol$ and by tautomerism, it changes into the acetaldehyde.
Tautomerism is structural isomerism where the only reallocation of a proton is taking place. This happens only for acetylene. Therefore, (A) is acetaldehyde.

Now the acetaldehyde is reacted ${{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}/{{\text{H}}^ + }$. We know that potassium dichromate in acidic medium is a strong oxidising agent. So, when aldehydes are oxidised corresponding carboxylic acids are formed from it.
$C{H_3} - CHO\xrightarrow{{{K_2}C{r_2}{O_7}/{H^ + }}}C{H_3}COOH$
Thus, we have (B) which is acetic acid.

Now, this acetic acid is reacted with calcium hydroxide. This happens in two steps.
$2C{H_3}COOH + Ca{\left( {OH} \right)_2} \to {\text{Ca}}{\left( {{\text{OCOC}}{{\text{H}}_3}} \right)_2} + 2{{\text{H}}_2}{\text{O}}$

The $- OH$ group from calcium hydroxide and ${H^ + }$ reacted to from water and we get the compound ${\text{Ca}}{\left( {{\text{OCOC}}{{\text{H}}_3}} \right)_2}$ which is our (C) product.

Lastly, when this product is heated, it results in the formation of Calcium carbonate and acetone. So, we will get:
$Ca{(OCOC{H_3})_2}\xrightarrow{\Delta }C{H_3} - CO - C{H_3} + CaC{O_3}$. This is the dry distillation process. And we get our product (D) which is acetone.

Therefore, the correct answer is option (C).

Note: We know that the dry distillation process is when solid materials are heated and converted to their gaseous state and then condensed to their liquid form. It is used here to separate calcium acetate into acetone and calcium carbonate. This is easily done because there is a difference in their boiling point.