Alkyne Questions in English

(D) The electrolysis of the aqueous solution of potassium salts of dicarboxylic acids is known as the Kolbe's electrolytic synthesis.
For the preparation of acetylene $(C_2H_2)$,potassium maleate or potassium fumarate is used.
The reaction is as follows:
$CHCOOK=CHCOOK + 2H_2O \xrightarrow{\text{electrolysis}} CH \equiv CH + 2CO_2 + 2KOH + H_2$
At the anode,acetylene and $CO_2$ are liberated.
Therefore,the compound '$A$' is potassium maleate.

(C) The hydrocarbon '$X$' reacts with sodamide $(NaNH_2)$,which indicates that '$X$' is a terminal alkyne.
Ozonolysis of a terminal alkyne followed by oxidative workup ($O_3$ followed by $Zn/H_2O_2$ or $H_2O_2$) leads to the cleavage of the triple bond to form a carboxylic acid and formic acid $(HCOOH)$.
For one of the products to be optically active,the carboxylic acid formed from the alkyl group must contain a chiral center.
Starting with $3-Methylpent-1-yne$ $(CH_3CH_2CH(CH_3)C \equiv CH)$:
$CH_3CH_2CH(CH_3)C \equiv CH \xrightarrow{O_3, H_2O_2} CH_3CH_2CH(CH_3)COOH + HCOOH$.
$2-Methylbutanoic acid$ $(CH_3CH_2CH(CH_3)COOH)$ contains a chiral carbon at the $C-2$ position,making it optically active.

(B) The acidity of alkynes depends on the stability of the conjugate base formed after the removal of the acidic proton.
Terminal alkynes ($I$ and $III$) have an acidic hydrogen atom attached to an $sp$-hybridized carbon,whereas internal alkynes $(II)$ do not have any acidic hydrogen.
Compound $(II)$ $(CH_3-C \equiv C-CH_3)$ has no acidic hydrogen,so it is the least acidic.
Between $(I)$ $(HC \equiv CH)$ and $(III)$ $(CH_3-C \equiv CH)$,the methyl group in $(III)$ exerts a $(+I)$ effect,which destabilizes the conjugate base (alkynide ion).
Therefore,the stability of the conjugate base follows the order: $HC \equiv C^- > CH_3-C \equiv C^-$.
Thus,the correct order of acidity is: $II < III < I$.

(D) The mixture of acetylene $(C_2H_2)$ and oxygen $(O_2)$ is used for oxy-acetylene welding.
This mixture produces a flame with a very high temperature,which is sufficient for cutting and welding metals effectively.

(A) The reaction of propyne $(CH_3-C \equiv CH)$ with two equivalents of hydrogen bromide $(HBr)$ follows Markovnikov's rule.
In the first step,$HBr$ adds to the triple bond to form $2-$bromopropene $(CH_3-C(Br)=CH_2)$.
In the second step,another molecule of $HBr$ adds to the double bond,again following Markovnikov's rule,to form $2,2-$dibromopropane $(CH_3-C(Br)_2-CH_3)$.
Therefore,the reaction $CH_3-C \equiv CH + 2 HBr \longrightarrow CH_3-C(Br)_2-CH_3$ is the correct one.

(D) The reaction of an alkyne with sodium in liquid ammonia $(Na/Liq. NH_3)$ is known as Birch reduction,which results in the stereoselective anti-addition of hydrogen to the triple bond to form a $trans$-alkene. The double bond present in the cyclohexene ring remains unaffected under these conditions. Therefore,the product is a $trans$-alkene. Thus,option $(D)$ is correct.

(D) The reaction of propyne $(CH_3-C \equiv CH)$ with excess $HBr$ follows Markovnikov's rule.
In the first step,$HBr$ adds across the triple bond to form $2-$bromopropene $(CH_3-C(Br)=CH_2)$.
In the second step,another molecule of $HBr$ adds to the alkene following Markovnikov's rule,where the hydrogen atom attaches to the carbon with more hydrogens and the bromine atom attaches to the more substituted carbon.
This results in the formation of $2,2-$dibromopropane $(CH_3-CBr_2-CH_3)$.

(D) $C_2D_2$ is deuteroacetylene and $CD_4$ is deuteromethane. These compounds are prepared by the deuterolysis of specific carbides.
$CaC_2$ is a calcium acetylide which reacts with $D_2O$ to produce $C_2D_2$ (deuteroacetylene).
$CaC_2 + 2D_2O \rightarrow C_2D_2 + Ca(OD)_2$
$Al_4C_3$ is an aluminum methanide which reacts with $D_2O$ to produce $CD_4$ (deuteromethane).
$Al_4C_3 + 12D_2O \rightarrow 3CD_4 + 4Al(OD)_3$
Thus,$X = CaC_2$ and $Y = Al_4C_3$.

(B) The chemical reaction is: $CaC_2 + 2 D_2 O \longrightarrow C_2 D_2 + Ca(OD)_2$.
Here,$X$ is $C_2 D_2$ (deuteroacetylene).
The structure of $C_2 D_2$ is $D-C \equiv C-D$.
Since each carbon atom is bonded to one deuterium atom by a single bond and to the other carbon atom by a triple bond,each carbon atom is $sp$ hybridized.

(D) $1$. The reaction of $CH_3CHBr_2$ with alcoholic $KOH$ followed by $NaNH_2$ leads to the formation of acetylene $(HC \equiv CH)$,which is represented as $A$ in the reaction sequence.
$2$. The hydration of acetylene in the presence of $Hg^{2+}$ and $H_2O$ at $333 \ K$ yields vinyl alcohol $(CH_2=CH-OH)$ as an intermediate $(B)$.
$3$. Vinyl alcohol is unstable and undergoes tautomerization to form the more stable keto form,which is acetaldehyde $(CH_3CHO)$,represented as $C$.

(C) The reaction of propyne $(C_3H_4)$ with water in the presence of $Hg^{2+} / H^{+}$ at $333 \ K$ is an electrophilic addition reaction.
First,the hydration of the alkyne leads to the formation of an unstable enol intermediate,$[X]$,which is prop$-1-$en$-2-$ol (an unsaturated alcohol).
This enol $[X]$ then undergoes tautomerisation to form a more stable keto form,$[Y]$,which is propanone (a ketone).
Therefore,$[X]$ is an unsaturated alcohol and $[Y]$ is a ketone.

(A) $1$. The Wacker process converts ethene $(\text{CH}_2=\text{CH}_2)$ to ethanal $(\text{CH}_3\text{CHO})$,so $B$ is $\text{CH}_3\text{CHO}$.
$2$. Ozonolysis of an alkene $A$ yields $B$ (ethanal). Since $A \xrightarrow{\text{O}_3} 2 \text{CH}_3\text{CHO}$,$A$ must be but$-2-$ene $(\text{CH}_3\text{CH}=\text{CHCH}_3)$.
$3$. The alkyne is reduced to $A$ (cis-but$-2-$ene) using Lindlar's catalyst. Therefore,the starting alkyne is but$-2-$yne $(\text{CH}_3\text{C} \equiv \text{CCH}_3)$.

(C) Acetylene $(HC \equiv CH)$ undergoes oxidation in the presence of alkaline potassium permanganate $(KMnO_4 / KOH)$ at $25^{\circ}C$ to form oxalic acid $(HOOC-COOH)$.
The reaction is represented as:
$HC \equiv CH + 4[O] \xrightarrow{KMnO_4 / KOH, 25^{\circ}C} HOOC-COOH$

(D) Acidity depends on the stability of the conjugate base formed after the removal of a proton. The stability of the conjugate base is increased by electron-withdrawing groups ($-I$ effect) and decreased by electron-donating groups ($+I$ effect).
$1$. In compound $II$,the $F_3C-$ group exerts a strong $-I$ effect,making it the most acidic.
$2$. In compound $I$,the terminal alkyne hydrogen is acidic due to the $sp$ hybridized carbon.
$3$. Compound $III$ is an internal alkyne with no acidic protons (no $C-H$ bond on the triple-bonded carbons),but if we consider the acidity of the $CH_3$ protons,they are less acidic than terminal alkynes.
$4$. Compound $IV$ is an alkene,which is the least acidic among the given options.
Thus,the correct order of acidity is $II > I > III > IV$.

(C) The molecular formula $C_6H_{10}$ corresponds to an alkyne with $6$ carbon atoms. The branched isomers are those that do not have a straight chain of $6$ carbons. The possible branched isomers are:
$1$. $3-\text{methyl}-1-\text{pentyne}$ $(CH_3-CH_2-CH(CH_3)-C \equiv CH)$
$2$. $4-\text{methyl}-1-\text{pentyne}$ $(CH_3-CH(CH_3)-CH_2-C \equiv CH)$
$3$. $3-\text{methyl}-2-\text{pentyne}$ $(CH_3-CH_2-C(CH_3)-C \equiv C-CH_3)$
$4$. $3,3-\text{dimethyl}-1-\text{butyne}$ $((CH_3)_2C(C \equiv CH)-CH_3)$
Thus,there are $4$ possible branched isomers.

(B) The given reaction is the reaction of ethylene dibromide with alcoholic $KOH$ to form acetylene.
In this reaction,two molecules of $HBr$ are removed from the vicinal dihalide,which is known as dehydrobromination (a type of dehydrohalogenation).

(C) $1$. The reaction of an alkyne with $H_2$ in the presence of Lindlar's catalyst (poisoned $Pd/CaCO_3$) is a syn-addition,which yields a (cis)-alkene as the major product $P$.
$2$. The reaction of an alkyne with $Na$ in liquid $NH_3$ (Birch reduction) is an anti-addition,which yields a (trans)-alkene as the major product $Q$.
$3$. Therefore,$P$ is (cis)-alkene and $Q$ is (trans)-alkene.

(D) The electrolysis of aqueous solutions of salts of dicarboxylic acids is known as Kolbe's electrolysis.
For the preparation of acetylene $(C_2H_2)$,potassium maleate or potassium fumarate is used.
The reaction is as follows:
$CHCOOK=CHCOOK + 2H_2O \xrightarrow{\text{electrolysis}} CH \equiv CH + 2CO_2 + 2KOH + H_2$
At the anode,acetylene and $CO_2$ are liberated.
Therefore,'$A$' is potassium maleate.

(A) The reduction of $2$-butyne with $Na / NH_3$ (liq.) (Birch reduction) yields the $E$-product (trans$-2-$butene) due to anti-addition of hydrogen.
Conversely,the catalytic hydrogenation of $2$-butyne using Lindlar's catalyst $(Pd / BaSO_4 + H_2)$ yields the $Z$-product (cis$-2-$butene) due to syn-addition of hydrogen.
Given the reaction scheme:
$Z$-product $\stackrel{Y}{\longleftarrow} 2$-butyne $\stackrel{X}{\longrightarrow} E$-product
$X$ must be $Na / NH_3$ (liq.) and $Y$ must be $Pd / BaSO_4 + H_2$.

(D) The general formula for an alkyne is $C_nH_{2n-2}$. For $C_6H_{10}$,$n=6$.
$1$-alkynes have the triple bond at the terminal position,represented as $R-C \equiv CH$.
For a $6$-carbon chain,the possible structures for $1$-alkynes are:
$1$. $CH_3-CH_2-CH_2-CH_2-C \equiv CH$ $(hex-1-yne)$
$2$. $CH_3-CH_2-CH(CH_3)-C \equiv CH$ $(3-methylpent-1-yne)$
$3$. $CH_3-CH(CH_3)-CH_2-C \equiv CH$ $(4-methylpent-1-yne)$
$4$. $(CH_3)_3C-C \equiv CH$ $(3,3-dimethylbut-1-yne)$
Thus,there are $4$ possible isomers for $1$-alkyne.

(C) An alkyne has acidic hydrogen if it is a terminal alkyne (triple bond at the end of the chain,$R-C \equiv CH$).
For the formula $C_6H_{10}$,the possible isomers are:
$1$. $Hex-1-yne$ $(CH_3CH_2CH_2CH_2C \equiv CH)$ - Terminal (Acidic)
$2$. $3-Methylpent-1-yne$ $(CH_3CH_2CH(CH_3)C \equiv CH)$ - Terminal (Acidic)
$3$. $4-Methylpent-1-yne$ $(CH_3CH(CH_3)CH_2C \equiv CH)$ - Terminal (Acidic)
$4$. $3,3-Dimethylbut-1-yne$ $((CH_3)_3CC \equiv CH)$ - Terminal (Acidic)
These $4$ isomers are terminal alkynes,so $x = 4$.
Non-terminal (internal) alkynes have no acidic hydrogens:
$1$. $Hex-2-yne$ $(CH_3CH_2CH_2C \equiv CCH_3)$
$2$. $Hex-3-yne$ $(CH_3CH_2C \equiv CCH_2CH_3)$
$3$. $4-Methylpent-2-yne$ $(CH_3CH(CH_3)C \equiv CCH_3)$
These $3$ isomers are internal alkynes,so $y = 3$.
Thus,$x = 4$ and $y = 3$.

(B) The starting material is $1,2-dibromopropane$ $(CH_3-CH(Br)-CH_2Br)$.
Step $1$: Treatment with $alc. KOH$ followed by $NaNH_2$ causes dehydrohalogenation to form propyne $(CH_3-C \equiv CH)$.
Step $2$: Hydration of propyne in the presence of $Hg^{2+}$ and $H^+$ (Kucherov's reaction) follows Markovnikov's rule.
Step $3$: The addition of water to the triple bond forms an enol intermediate $(CH_3-C(OH)=CH_2)$,which tautomerizes to form propanone $(CH_3-CO-CH_3)$.

(B) When propyne $(CH_3-C \equiv CH)$ is passed through a red hot iron tube at $873 \ K$,it undergoes cyclic trimerization to form an aromatic compound called mesitylene ($1,3,5$-trimethylbenzene).
The reaction is as follows:
$3CH_3-C \equiv CH \xrightarrow{\text{Red hot Fe, } 873 \ K} C_9H_{12}$ (Mesitylene).
The molecular formula of mesitylene is $C_9H_{12}$.

(D) The reaction of an alkyne with excess $Br_2$ in $CCl_4$ is an electrophilic addition reaction.
During this process,both $\pi$-bonds of the triple bond are broken,and four new $\sigma$-bonds are formed with bromine atoms.
For hex$-1-$yne $(CH_3(CH_2)_3C \equiv CH)$,the addition occurs at the $C-1$ and $C-2$ positions.
The final product is $1,1,2,2-$tetrabromohexane.
$CH_3(CH_2)_3C \equiv CH + 2Br_2 \xrightarrow{CCl_4} CH_3(CH_2)_3C(Br)_2-CH(Br)_2$

(A) The reaction of an internal alkyne with an alkali metal (like $Na$ or $Li$) in liquid $NH_3$ is known as a dissolving metal reduction or Birch-type reduction of alkynes.
This reaction proceeds via a radical anion intermediate and stereoselectively yields the $trans$-alkene as the major product due to the greater stability of the $trans$-vinyl radical intermediate.
Therefore,the reduction of $but-2-yne$ $(CH_3-C \equiv C-CH_3)$ with $Na/liquid \ NH_3$ produces $trans-but-2-ene$.

(D) The reaction of an internal alkyne like $2-$pentyne with sodium in liquid ammonia $(Na/Liq. NH_3)$ is a Birch reduction,which stereoselectively produces a $trans-$alkene.
$CH_3-C \equiv C-CH_2-CH_3 \xrightarrow{Na/Liq. NH_3} CH_3-CH=CH-CH_2-CH_3$ (trans$-2-$pentene).
Therefore,compound $A$ is trans$-2-$pentene.

(C) The reaction of propyne $(CH_3-C \equiv CH)$ with $2$ equivalents of $HBr$ proceeds via Markovnikov's addition.
In the first step,$HBr$ adds to the triple bond to form $CH_3-CBr=CH_2$ ($2$-bromopropene).
In the second step,another molecule of $HBr$ adds to the double bond following Markovnikov's rule,where the hydrogen atom attaches to the carbon with more hydrogens,resulting in the geminal dihalide $CH_3-CBr_2-CH_3$ ($2$,$2$-dibromopropane).

(D) The reaction of calcium carbide $(CaC_2)$ with heavy water $(D_2O)$ is an acid-base type reaction where the carbide ion $(C_2^{2-})$ reacts with $D^+$ ions from $D_2O$.
$CaC_2 + 2D_2O \rightarrow Ca(OD)_2 + C_2D_2$
Thus,the product "$P$" is $C_2D_2$ (deuteroacetylene).

(B) Calcium carbide $(CaC_2)$ is an ionic carbide containing the acetylide ion $[C \equiv C]^{2-}$.
When it reacts with heavy water $(D_2O)$,it undergoes hydrolysis to form deuterated acetylene and calcium deuteroxide.
The chemical reaction is:
$CaC_2 + 2D_2O \rightarrow C_2D_2 + Ca(OD)_2$
From the reaction,the products formed are $C_2D_2$ (labeled as $II$) and $Ca(OD)_2$ (labeled as $III$).

(A) The given reaction sequence is:
$1$. $[P]$ is ethene $(CH_{2}=CH_{2})$. Reaction with $Br_{2}$ gives $1,2-dibromoethane$ $(CH_{2}Br-CH_{2}Br)$.
$2$. $1,2-dibromoethane$ reacts with $NaNH_{2}/NH_{3}$ (dehydrohalogenation) to form ethyne $(HC \equiv CH)$ as $[Q]$.
$3$. Ethyne $[Q]$ reacts with $20 \% H_{2}SO_{4}$ and $Hg^{2+}$ (Kucherov reaction) to form ethanal $(CH_{3}CHO)$ as $[R]$.
$4$. Ethanal $[R]$ undergoes Clemmensen reduction $(Zn-Hg/HCl)$ to form ethane $(CH_{3}CH_{3})$ as $[S]$.
Thus,$[P]$ = ethene,$[Q]$ = ethyne,$[R]$ = ethanal,$[S]$ = ethane.

(A) $1$. Compound $M$ reacts with ammoniacal $AgNO_3$ to give a white precipitate,which indicates that $M$ is a terminal alkyne $(R-C \equiv CH)$.
$2$. $M$ undergoes hydrogenation with $H_2$ and Lindlar catalyst to form $N$,which is an alkene.
$3$. $N$ undergoes ozonolysis to give $O$ and $P$. $O$ (benzaldehyde,$Ph-CHO$) reacts with $(CH_3CO)_2O$ and $CH_3COONa$ (Perkin's condensation) to form cinnamic acid $(Ph-CH=CH-COOH)$.
$4$. Based on the reaction sequence,$M$ must be phenylacetylene $(Ph-C \equiv CH)$.
$5$. $Ph-C \equiv CH + H_2 \xrightarrow{Lindlar} Ph-CH=CH_2$ $(N)$.
$6$. $Ph-CH=CH_2 \xrightarrow{O_3} Ph-CHO$ $(O)$ + $HCHO$ $(P)$.
$7$. Thus,$M$ is $Ph-C \equiv CH$.

(D) The starting material is $but-2-yne$ (or $CH_{3}-C \equiv C-CH_{3}$),but the image shows $prop-1-yne$ $(CH_{3}-C \equiv CH)$ being converted to $butan-2-one$ $(CH_{3}-CO-CH_{2}-CH_{3})$.
Step $1$: $CH_{3}-C \equiv CH + NaNH_{2} \rightarrow CH_{3}-C \equiv C^-Na^+ + NH_{3}$.
Step $2$: $CH_{3}-C \equiv C^-Na^+ + CH_{3}I \rightarrow CH_{3}-C \equiv C-CH_{3} + NaI$.
Step $3$: $CH_{3}-C \equiv C-CH_{3} + H_{2}O \xrightarrow{HgSO_{4} / dil. H_{2}SO_{4}} CH_{3}-C(OH)=CH-CH_{3}$ (enol form).
Step $4$: The enol form undergoes tautomerization to form $CH_{3}-CO-CH_{2}-CH_{3}$ $(butan-2-one)$.
Thus,the correct set of reagents is $NaNH_{2} / CH_{3}I ; HgSO_{4} / dil. \ H_{2}SO_{4}$.

(B) The reaction sequence involves the dehydrohalogenation of a vicinal dibromide using $NaNH_2$ to form an alkyne. The first step with $NaNH_2$ removes two molecules of $HBr$ to produce $hex-2-yne$ $(CH_3-C \equiv C-CH_2-CH_2-CH_3)$.
The second step involves the reduction of the alkyne using $Na$ in liquid $NH_3$ (Birch reduction),which stereoselectively reduces the internal alkyne to a $trans-alkene$.
Therefore,the final product $X$ is $trans-hex-2-ene$.

(D) $CH_{3}C \equiv CMgBr$ is a Grignard reagent derived from a terminal alkyne.
It can be prepared by the acid-base reaction between propyne $(CH_{3}C \equiv CH)$ and methylmagnesium bromide $(CH_{3}MgBr)$.
The terminal hydrogen of the alkyne is acidic and reacts with the basic Grignard reagent to form the alkynyl magnesium bromide and methane gas.
The chemical equation is:
$CH_{3}C \equiv CH + CH_{3}MgBr \longrightarrow CH_{3}C \equiv CMgBr + CH_{4}$

(D) The isomerisation of $1-$butyne $(CH_3-CH_2-C \equiv CH)$ to $2-$butyne $(CH_3-C \equiv C-CH_3)$ is a base-catalyzed reaction.
When $1-$butyne is treated with ethanolic $KOH$ at high temperatures,it undergoes base-catalyzed isomerisation to form the more stable internal alkyne,$2-$butyne.
The reaction is as follows:
$CH_3-CH_2-C \equiv CH \xrightarrow{\text{ethanolic } KOH, \Delta} CH_3-C \equiv C-CH_3$

(C) Acetylene $(HC \equiv CH)$ is a terminal alkyne with acidic hydrogen atoms.
When passed through ammoniacal cuprous chloride $(Cu_2Cl_2)$,it forms a red precipitate of copper$(I)$ acetylide:
$HC \equiv CH + Cu_2Cl_2 \rightarrow CuC \equiv CCu \downarrow (\text{red ppt}) + 2HCl$.
Ethene $(CH_2=CH_2)$ does not react with ammoniacal $Cu_2Cl_2$ and passes through unchanged.
Therefore,the mixture can be separated using ammoniacal $Cu_2Cl_2$.

(A) Cyclo$[18]$ carbon $(C_{18})$ is a cyclic allotrope of carbon consisting of $18$ carbon atoms.
In this structure,the carbon atoms are connected by alternating single and triple bonds.
Since there are $18$ carbon atoms in the ring,and each triple bond involves two carbon atoms while each single bond connects two adjacent triple-bonded units,the structure consists of $9$ triple bonds and $9$ single bonds.
Therefore,the total number of triple bonds is $9$.

(B) Calcium carbide is calcium acetylide,$CaC_2$,with the structure $[Ca]^{2+} [:C \equiv C:]^{2-}$.
In the acetylide ion $[:C \equiv C:]^{2-}$,there is a triple bond between the two carbon atoms.
$A$ triple bond consists of one sigma $(\sigma)$ bond and two pi $(\pi)$ bonds.
Therefore,the correct option is $B$.

(B) The hydrogenation of $But-2-yne$ with $Pd/C$ (Lindlar's catalyst or similar) gives $cis-But-2-ene$ $(X)$.
The reduction of $But-2-yne$ with $Na / \text{liq. } NH_3$ (Birch reduction) gives $trans-But-2-ene$ $(Y)$.
$A$. $X$ and $Y$ are geometric isomers (stereoisomers). This statement is correct.
$B$. $X$ $(cis-But-2-ene)$ has a non-zero dipole moment due to the polarity of the $C-CH_3$ bonds. $Y$ $(trans-But-2-ene)$ has a dipole moment of zero. Thus,statement $B$ is incorrect.
$C$. $cis-But-2-ene$ $(X)$ is more polar than $trans-But-2-ene$ $(Y)$,so $X$ has a higher boiling point than $Y$. This statement is correct.
$D$. Both $X$ and $Y$ are isomers of $But-2-ene$. Ozonolysis of both $cis-But-2-ene$ and $trans-But-2-ene$ yields the same product,$acetaldehyde$ $(CH_3CHO)$. Thus,statement $D$ is incorrect.
The incorrect statements are $B$ and $D$.

(B) Step $1$: Dehydrohalogenation of $1,2-$dibromopentane with $2NaNH_2$ gives pent$-1-$yne $(X = CH_3CH_2CH_2C \equiv CH)$.
Step $2$: Treatment of pent$-1-$yne with $NaNH_2$ forms the sodium acetylide salt $(CH_3CH_2CH_2C \equiv C^-Na^+)$.
Step $3$: The acetylide ion acts as a nucleophile and attacks isopropyl bromide $((CH_3)_2CHBr)$ via an $S_N2$ mechanism.
$CH_3CH_2CH_2C \equiv C^- + (CH_3)_2CHBr \rightarrow CH_3CH_2CH_2C \equiv C-CH(CH_3)_2$.
The resulting product is $5-$methylhex$-2-$yne.

(B) $1$. Styrene $(C_6H_5CH=CH_2)$ reacts with $Br_2/CHCl_3$ to form styrene dibromide $(C_6H_5CH(Br)CH_2Br)$.
$2$. Treatment with excess $NaNH_2$ causes dehydrohalogenation to form phenylacetylene $(C_6H_5C\equiv CH)$.
$3$. Reaction with $CH_3I$ (after deprotonation by $NaNH_2$) yields $1$-phenylprop-$1$-yne $(C_6H_5C\equiv CCH_3)$.
$4$. Reduction with $H_2, Na/NH_3(l)$ (Birch-like reduction of alkyne) yields $(E)$-$1$-phenylprop-$1$-ene $(C_6H_5CH=CHCH_3)$.
$5$. The molecular formula of the product $(Y)$ is $C_9H_{10}$.
$6$. Molar mass $= (9 \times 12) + (10 \times 1) = 108 + 10 = 118 \text{ g mol}^{-1}$.