$CH_3CH_2CH_2I$ $\xrightarrow{alc. KOH} A$ $\xrightarrow{Br_2} B$ $\xrightarrow{NaNH_2 / NH_3} C$. In this reaction,$C$ is:

When calcium carbide is reacted with heavy water,which of the following product$(s)$ will be formed?
$I$. $CD_4$
$II$. $C_2D_2$
$III$. $Ca(OD)_2$
$IV$. $Ca_2 \cdot D_2O$

Acetylene can be prepared from

$A$ gas decolourises bromine in $CCl_4$ and forms a precipitate with ammoniacal silver nitrate. The gas is

Schemes $1$ and $2$ describe the sequential transformation of alkynes $M$ and $N$. Consider only the major products formed in each step for both the schemes.
$1.$ The product $X$ is:
$2.$ The correct statement with respect to product $Y$ is:
$(A)$ It gives a positive Tollens test and is a functional isomer of $X$.
$(B)$ It gives a positive Tollens test and is a geometrical isomer of $X$.
$(C)$ It gives a positive iodoform test and is a functional isomer of $X$.
$(D)$ It gives a positive iodoform test and is a geometrical isomer of $X$.

In order to complete the reaction $1-Pentyne$ $\xrightarrow{a} 4-Octyne$ $\xrightarrow{b} cis-4-Octene$,$a$ and $b$ will be:
$A) NaNH_2; CH_3CH_2CH_2Br : H_2, (1 \ mole) \text{ Lindlar catalyst}$
$B) NaNH_2; CH_3CH_2CH_2Br : H_2, (2 \ moles) \text{ Pd or Ni}$
$C) NaNH_2; CH_3CH_2CH_2Br : H_2, (1 \ mole) \text{ Pd or Ni}$
$D) NaNH_2; CH_3CH_2CH_2Br : BH_3, H_2O_2, OH^-$