Alkene Questions in English

(C) The reaction of an alkyl bromide with sodium ethoxide $(C_2H_5ONa)$ in ethanol $(C_2H_5OH)$ is a dehydrohalogenation reaction (elimination reaction).
For the alkyl bromide to produce a single alkene,it must have only one possible $\beta-$hydrogen elimination path.
$1-$bromo$-2-$methylbutane $(CH_3-CH_2-CH(CH_3)-CH_2Br)$ undergoes elimination to form $2-$methylbut$-1-$ene $(CH_3-CH_2-C(CH_3)=CH_2)$.
Hydrogenation of $2-$methylbut$-1-$ene $(CH_3-CH_2-C(CH_3)=CH_2 + H_2 \rightarrow CH_3-CH_2-CH(CH_3)-CH_3)$ yields $2-$methylbutane.
Therefore,the correct alkyl bromide is $1-$bromo$-2-$methylbutane.

(C) Dehydrohalogenation of $1, 2$-dibromocyclohexane involves the removal of two molecules of $HBr$ in the presence of a strong base (like alcoholic $KOH$).
Initially,$1, 2$-dibromocyclohexane undergoes elimination to form bromocyclohexene.
Further elimination of the second $HBr$ molecule leads to the formation of $1, 3$-cyclohexadiene as the major product due to the conjugation of the double bonds,which provides extra stability.

(B) The conversion of propene to $1-propanol$ is an anti-Markovnikov hydration reaction,which is achieved via hydroboration-oxidation.
Step $1$: Hydroboration: $3CH_3-CH=CH_2 + \frac{1}{2}B_2H_6 \rightarrow (CH_3-CH_2-CH_2)_3B$
Step $2$: Oxidation: $(CH_3-CH_2-CH_2)_3B + 3H_2O_2 \xrightarrow{OH^-} 3CH_3-CH_2-CH_2-OH + B(OH)_3$
Thus,the reagents $B_2H_6$ and alkaline $H_2O_2$ are ideal for this conversion.

(B) The reaction given is the hydroboration-oxidation of ethene.
$1$. In the first step,$B_2H_6$ (diborane) adds across the double bond of ethene $(CH_2=CH_2)$ to form a trialkylborane intermediate.
$2$. In the second step,the trialkylborane is oxidized by hydrogen peroxide $(H_2O_2)$ in the presence of an aqueous base $(NaOH)$.
$3$. This process results in the anti-Markovnikov addition of water across the double bond,yielding ethanol $(CH_3CH_2OH)$ as the final product.

(A) The hydration of alkenes in the presence of $H_2SO_4$ follows the Markownikoff rule,where the hydroxyl group $(-OH)$ attaches to the more substituted carbon atom.
$(i)$ For $CH_3-CH=C(CH_3)_2$ ($2$-methylbut$-2-$ene):
The double bond is between $C2$ and $C3$. The $C3$ atom is more substituted than $C2$. Thus,the $-OH$ group attaches to $C3$,yielding $CH_3-CH_2-C(OH)(CH_3)_2$ ($2$-methylbutan$-2-$ol).
(ii) For $CH_3-CH=CH_2$ (propene):
The double bond is between $C1$ and $C2$. The $C2$ atom is more substituted than $C1$. Thus,the $-OH$ group attaches to $C2$,yielding $CH_3-CH(OH)-CH_3$ (propan$-2-$ol).
Therefore,the correct products are:
$(i)$ $CH_3-CH_2-C(OH)(CH_3)_2$
(ii) $CH_3-CH(OH)-CH_3$

(D) Alkenes can be converted into alcohols or diols through various hydration processes.
$1$. Alkaline $KMnO_{4}$ (Baeyer's reagent) causes syn-hydroxylation of alkenes to form vicinal diols (glycols): $CH_{2}=CH_{2} + H_{2}O + [O] \xrightarrow{\text{alk. } KMnO_{4}} CH_{2}(OH)-CH_{2}(OH)$.
$2$. Hydration of alkenes using conc. $H_{2}SO_{4}$ followed by hydrolysis with water vapours yields alcohols: $CH_{2}=CH_{2} + H_{2}O \xrightarrow{\text{conc. } H_{2}SO_{4}} CH_{3}-CH_{2}-OH$.
Since both processes result in the formation of hydroxyl-containing compounds,the correct answer is $(d)$.

(B) Acid-catalyzed hydration of alkenes follows Markovnikov's rule,where the $-OH$ group adds to the more substituted carbon atom of the $C=C$ double bond.
For ethene $(CH_2=CH_2)$,hydration yields primary alcohol (ethanol).
For other alkenes (e.g.,prop$-1-$ene,$2$-methylprop$-1-$ene),the addition of water results in the formation of secondary or tertiary alcohols,as shown in the reaction schemes:
$1$. Prop$-1-$ene $(CH_3-CH=CH_2)$ $\xrightarrow{H_2O/H^+}$ Propan$-2-$ol (Secondary alcohol).
$2$. $2-$Methylprop$-1-$ene $((CH_3)_2C=CH_2)$ $\xrightarrow{H_2O/H^+}$ $2-$Methylpropan$-2-$ol (Tertiary alcohol).

(C) When glycerol $(CH_2OH-CHOH-CH_2OH)$ is treated with excess hydroiodic acid $(HI)$,it undergoes a series of reactions:
$1.$ First,it forms $1,2,3-triiodopropane$,which is unstable and loses $I_2$ to form allyl iodide $(CH_2=CH-CH_2I)$.
$2.$ Allyl iodide reacts with another molecule of $HI$ to form $1,2-diiodopropane$,which is also unstable and loses $I_2$ to form propene $(CH_3-CH=CH_2)$.
$3.$ Thus,the final stable product obtained under these conditions is propene.

(C) The dehydration of ethanol $(CH_{3}CH_{2}OH)$ in the presence of concentrated sulfuric acid $(H_{2}SO_{4})$ at $170 \ ^{o}C$ leads to the formation of ethylene $(CH_{2}=CH_{2})$ and water.
$CH_{3}CH_{2}OH \xrightarrow[170 \ ^{o}C]{\text{Conc. } H_{2}SO_{4}} CH_{2}=CH_{2} + H_{2}O$

(D) The reaction of ethanol with concentrated $H_2SO_4$ at $170\,^{\circ}C$ is a dehydration reaction.
In this process,ethanol undergoes elimination of a water molecule to form ethene.
The chemical equation is:
$C_2H_5OH \xrightarrow{\text{Conc. } H_2SO_4, 170\,^{\circ}C} C_2H_4 + H_2O$
Therefore,the correct option is $D$.

(A) At $260\,^\circ C$,glycerol reacts with oxalic acid to produce allyl alcohol.
The reaction proceeds through the formation of glycerol dioxalate,which then undergoes decarboxylation and elimination to yield allyl alcohol $(CH_2=CH-CH_2OH)$.

(B) Alkenes react with bromine in $CCl_4$ to undergo electrophilic addition,resulting in the decolourisation of the reddish-brown bromine solution. $CH_2=CH_2 + Br_2/CCl_4 \rightarrow CH_2Br-CH_2Br$ (colourless).
Alcohols do not react with bromine in $CCl_4$ under these conditions.
Therefore,decolourising with bromine in $CCl_4$ is a standard test to distinguish alkenes from alcohols.

(A) When glycerol is heated with oxalic acid at $380 \ K$,it produces formic acid and carbon dioxide.
However,when the reaction is carried out at a higher temperature of $530 \ K$,glycerol reacts with oxalic acid to form allyl alcohol $(CH_2=CH-CH_2OH)$.
Therefore,the correct product at $530 \ K$ is allyl alcohol.

(C) Ethylene glycol $(HO-CH_2-CH_2-OH)$ when heated with anhydrous $ZnCl_2$ undergoes dehydration to form vinyl alcohol $(CH_2=CH-OH)$,which tautomerizes to form acetaldehyde $(CH_3-CHO)$.
$HO-CH_2-CH_2-OH$ $\xrightarrow{ZnCl_2, \Delta} [CH_2=CH-OH]$ $\rightarrow CH_3-CHO$

(C) $1.$ The dehydration of $2-methylbutanol$ $(CH_3CH_2CH(CH_3)CH_2OH)$ with conc. $H_2SO_4$ proceeds via the formation of a carbocation intermediate.
$2.$ Initially,a primary carbocation is formed,which undergoes a $1,2-hydride$ shift to form a more stable tertiary carbocation $(CH_3CH_2C^+(CH_3)CH_3)$.
$3.$ According to Saytzeff's rule,the more substituted alkene is the major product.
$4.$ Elimination of a proton from the tertiary carbocation yields $2-methylbut-2-ene$ $(CH_3CH=C(CH_3)CH_3)$ as the major product.

(C) The reaction of ethanol $(CH_3CH_2OH)$ with $Cu$ at $573 \ K$ $(\Delta)$ is a dehydrogenation reaction that produces an aldehyde (alkanal),$CH_3CHO$ $(A)$.
The reaction of ethanol $(CH_3CH_2OH)$ with $Al_2O_3$ at $623 \ K$ $(\Delta)$ is a dehydration reaction that produces an alkene,$CH_2=CH_2$ $(B)$.
Therefore,$A$ is an alkanal and $B$ is an alkene.

(D) The reaction of ethanol $(CH_3CH_2OH)$ with hot concentrated $H_2SO_4$ at $170 \ ^oC$ leads to dehydration,where a molecule of water is eliminated to form ethene.
The reaction is: $CH_3CH_2OH \xrightarrow{Conc. H_2SO_4, 170 \ ^oC} CH_2=CH_2 + H_2O$.

(C) $Ethylene$ $(CH_2=CH_2)$ reacts with Baeyer's reagent (cold,dilute,alkaline $KMnO_4$) to undergo hydroxylation,forming $Ethylene$ $glycol$ $(HO-CH_2-CH_2-OH)$.
$CH_2=CH_2 + H_2O + [O] \xrightarrow{\text{cold alk. } KMnO_4} HO-CH_2-CH_2-OH$

(C) The preparation of $Propan-1-ol$ from $propene$ is an example of anti-Markovnikov hydration.
This is achieved via hydroboration-oxidation using $B_2H_6$ followed by $H_2O_2/OH^-$.
The reaction is:
$CH_3-CH=CH_2 + B_2H_6 \xrightarrow{H_2O_2/OH^-} CH_3-CH_2-CH_2OH$.

(B) Ozonolysis of alkenes involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
For the formation of acetone $(CH_3COCH_3)$,the alkene must have two methyl groups attached to each of the doubly bonded carbon atoms.
In the reaction: $(CH_3)_2C=C(CH_3)_2 \xrightarrow{O_3, Zn/H_2O} 2CH_3COCH_3$.
Thus,$2,3-dimethylbut-2-ene$ gives two molecules of acetone upon ozonolysis.

(C) The ozonolysis of $but-1-ene$ $(CH_3CH_2CH=CH_2)$ involves the addition of ozone to form an ozonide intermediate.
Upon hydrolysis,the ozonide breaks down to form carbonyl compounds. The cleavage occurs at the double bond site:
$CH_3CH_2CH=CH_2$ $\xrightarrow{O_3} \text{Ozonide}$ $\xrightarrow{H_2O/Zn} CH_3CH_2CHO + HCHO$
Here,$CH_3CH_2CHO$ is propionaldehyde (propanal) and $HCHO$ is formaldehyde (methanal). Therefore,the hydrolysis of the ozonide of $but-1-ene$ yields propionaldehyde and formaldehyde.

(B) The reaction of ethene $(CH_2=CH_2)$ with ozone $(O_3)$ forms an ozonide intermediate.
Upon reductive hydrolysis with $Zn/H_2O$,the ozonide undergoes cleavage to produce two molecules of formaldehyde $(HCHO)$.

(A) Reductive ozonolysis of an alkene involves the cleavage of the $C=C$ double bond and the addition of oxygen atoms to each carbon to form carbonyl compounds.
For $1-$methylcyclohexene,the $C=C$ bond breaks,opening the ring to form a linear chain.
The carbon atom originally attached to the methyl group becomes a ketone $(C=O)$,and the other carbon of the double bond becomes an aldehyde $(-CHO)$.
The resulting product is a $7-$carbon chain with a ketone group at position $6$ and an aldehyde group at position $1$,which is $6-$oxoheptanal.

(A) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond and the addition of oxygen atoms to each carbon atom to form carbonyl compounds.
Given products are $2,2$-dimethylpropanal $(CH_3)_3C-CHO$ and butan$-2$-one $(CH_3-CO-CH_2CH_3)$.
To find the original alkene,we join the carbonyl carbons by removing the oxygen atoms and forming a double bond between them:
$(CH_3)_3C-CH=O + O=C(CH_3)CH_2CH_3 \rightarrow (CH_3)_3C-CH=C(CH_3)CH_2CH_3$.
The resulting structure is $2,2,4$-trimethylhex$-3$-ene.

(A) This is a $Wittig$ reaction where a phosphorus ylide reacts with a ketone $(2-\text{Butanone})$ to form an alkene.
$CH_3-CH_2-CH_2-CH=PPh_3 + CH_3-CO-CH_2-CH_3 \rightarrow CH_3-CH_2-CH_2-CH=C(CH_3)-CH_2-CH_3 + Ph_3P=O$
The resulting product is $3-\text{Methylhept-3-ene}$.

(D) Vigorous oxidation of alkenes with hot alkaline $KMnO_4$ leads to the cleavage of the $C=C$ double bond.
For the alkene $(CH_3)_2C = CH - CH_2CH_2CH_3$ ($2$-methylhex$-2-$ene),the cleavage occurs at the double bond:
$1$. The $(CH_3)_2C=$ part is oxidized to acetone,$(CH_3)_2C=O$.
$2$. The $=CH-CH_2CH_2CH_3$ part is oxidized to a carboxylic acid,$CH_3CH_2CH_2COOH$ (butanoic acid).
Looking at the options,option $D$ shows the formation of acetone and propanoic acid,which is a common error in such questions if the carbon chain length is miscounted. However,based on the structure $(CH_3)_2C = CH - CH_2CH_2CH_3$,the products are acetone and butanoic acid. Given the provided options,option $D$ is the closest structural match representing the cleavage products $R_2C=O$ and $R'COOH$.

(B) $N$-bromosuccinimide $(NBS)$ is a reagent used for allylic bromination. When propene $(CH_3 - CH = CH_2)$ reacts with $NBS$, it undergoes free radical substitution at the allylic position (the carbon atom adjacent to the double bond).
The reaction is as follows:
$CH_3 - CH = CH_2 + NBS \rightarrow BrCH_2 - CH = CH_2 + \text{succinimide}$
The product formed is allyl bromide $(BrCH_2 - CH = CH_2)$.

(B) In the presence of peroxide,the addition of $HBr$ to an unsymmetrical alkene follows the anti-Markovnikov rule (Kharasch effect).
The bromine atom attaches to the terminal carbon atom of the double bond.
$CH_2 = CH - (CH_2)_5COOH + HBr \xrightarrow{\text{Peroxide}} BrCH_2 - CH_2 - (CH_2)_5COOH$,which is $BrCH_2 - (CH_2)_6COOH$.

(C) The reaction of an alkene with $HBr$ in the presence of a peroxide follows the anti-Markovnikov addition mechanism (Kharasch effect or peroxide effect).
In this reaction,the $Br^\bullet$ radical adds to the carbon atom of the double bond that has more hydrogen atoms,resulting in the formation of the less substituted radical.
For the reactant $CH_2=CH(CH_2)_8COOH$,the terminal carbon $(CH_2=)$ has two hydrogen atoms,while the internal carbon $(-CH=)$ has one.
Therefore,the $Br$ atom attaches to the terminal $CH_2$ group,and the $H$ atom attaches to the internal $CH$ group.
The product formed is $BrCH_2CH_2(CH_2)_8COOH$.

(C) . Lactic acid $(CH_3-CH(OH)-COOH)$ on heating with concentrated $H_2SO_4$ undergoes dehydration to form acrylic acid $(CH_2=CH-COOH)$.
$CH_3-CH(OH)-COOH \xrightarrow{\text{conc. } H_2SO_4, \Delta} CH_2=CH-COOH + H_2O$

(A) The decolorization of $Br_2/CCl_4$ is a characteristic test for the presence of an unsaturated carbon-carbon double bond $(C=C)$.
Cinnamic acid $(C_6H_5CH=CHCOOH)$ contains a carbon-carbon double bond in its side chain.
It reacts with bromine in carbon tetrachloride via electrophilic addition to form dibromocinnamic acid,which is colorless.
$C_6H_5CH=CHCOOH + Br_2 \xrightarrow{CCl_4} C_6H_5CH(Br)-CH(Br)COOH$
Benzoic acid,$o-$phthalic acid,and acetophenone do not contain $C=C$ bonds that undergo this addition reaction under these conditions.

(B) $LiAlH_4$ is a strong reducing agent that reduces carboxylic acids to primary alcohols but does not reduce isolated carbon-carbon double bonds.
Therefore,$CH_2=CH-COOH \xrightarrow{LiAlH_4} CH_2=CH-CH_2OH$.

(C) Mustard gas,also known as $1,1'$-thiobis($2$-chloroethane),is synthesized by the reaction of ethylene $(CH_2=CH_2)$ with sulfur dichloride $(SCl_2)$.
The chemical equation is: $2CH_2=CH_2 + SCl_2 \rightarrow (ClCH_2CH_2)_2S$.

(C) The test for unsaturation in alkenes is performed using Baeyer's reagent,which is cold dilute alkaline $KMnO_4$ solution.
During this reaction,the purple color of $KMnO_4$ disappears,and a brown precipitate of $MnO_2$ is formed,indicating the presence of a double bond.
This reaction is known as the hydroxylation of alkenes to form vicinal glycols.

(B) Baeyer's reagent is defined as a cold,dilute,alkaline solution of potassium permanganate $(KMnO_4)$.
It is used as an oxidizing agent to test for the presence of unsaturation (double or triple bonds) in organic compounds.
When an alkene reacts with Baeyer's reagent,the purple color of $KMnO_4$ disappears,and a brown precipitate of $MnO_2$ is formed,indicating the formation of a vicinal diol.
Therefore,the correct composition is $KMnO_4 + KOH$.

(A) Step $1$: $CH_3CH_2CH_2Br + aq. NaOH \rightarrow CH_3CH_2CH_2OH$ ($X$ is propan$-1-$ol).
Step $2$: $CH_3CH_2CH_2OH \xrightarrow{Al_2O_3, \text{Heat}} CH_3CH=CH_2$ ($Y$ is propene).
Step $3$: $CH_3CH=CH_2 + HOCl \rightarrow CH_3CH(OH)CH_2Cl$ (major product) and $CH_3CH(Cl)CH_2OH$ (minor product).
According to Markovnikov's rule,the electrophile $Cl^+$ attacks the double bond to form the more stable carbocation,followed by the attack of $OH^-$. The major product is $CH_3CH(OH)CH_2Cl$.

(B) The peroxide effect (Kharasch effect) is only applicable to the addition of $HBr$ to unsymmetrical alkenes.
For $HCl$,the addition follows Markovnikov's rule even in the presence of peroxide.
In the addition of $HCl$ to propene $(CH_3-CH=CH_2)$,the electrophile $H^+$ attacks the double bond to form the more stable secondary carbocation,$CH_3-CH^+-CH_3$.
Therefore,the intermediate formed is the secondary carbocation,$CH_3-CH^+-CH_3$.

(A) The addition of $HBr$ to propene follows an electrophilic addition mechanism.
In the first step,the electrophile $H^{+}$ attacks the double bond of the propene molecule to form a carbocation intermediate.

(C) The debromination of $meso-2,3-dibromobutane$ occurs via an $E2$ elimination mechanism using zinc dust in ethanol.
In the $meso$ isomer,the two bromine atoms are in an anti-periplanar conformation when the molecule is in its eclipsed or gauche-like state,but specifically,the $meso$ form allows for the anti-elimination of the two bromine atoms to yield the $cis-2-butene$ isomer.
Therefore,the major product is $cis-2-butene$.

(C) In ethene $(CH_2=CH_2)$, the carbon atoms are $sp^2$ hybridized.
The $C-H$ bond length in ethene is approximately $108 \ pm$ to $110 \ pm$.
Among the given options, $110 \ pm$ is the most accurate value for the $C-H$ bond length in ethene.

(B) The reaction of propene $(CH_3-CH=CH_2)$ with $HCl$ follows the Markovnikov addition rule.
Peroxide effect (Kharasch effect) is only applicable to $HBr$ and not to $HCl$ or $HI$.
Therefore,the addition of $HCl$ to propene in the presence of peroxide still follows the Markovnikov rule,where the $Cl^-$ ion attaches to the more substituted carbon atom.
Thus,the major product is $2$-chloropropane $(CH_3-CHCl-CH_3)$.

(C) The reaction involves the anti-addition of $Br_2$ to trans-but$-2-$ene.
Trans-alkene + $Br_2$ (anti-addition) yields a meso compound.
Starting with trans-but$-2-$ene,the anti-addition of bromine results in the formation of meso$-2,3-$dibromobutane.
Looking at the Fischer projections provided,option $C$ represents the meso$-2,3-$dibromobutane structure,which has a plane of symmetry.

(B) The reaction of $1$-butene $(CH_3-CH_2-CH=CH_2)$ with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition rule (Kharasch effect or peroxide effect).
According to this rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom of the double bond that has the greater number of hydrogen atoms.
Therefore,the reaction is: $CH_3-CH_2-CH=CH_2 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2-CH_2Br$.
The product formed is $1$-bromobutane.

(A) The addition of $HCl$ to $1,3$-butadiene is a classic example of kinetic versus thermodynamic control.
At low temperatures ($0^{\circ}C$ or below),the reaction is under kinetic control,which favors the formation of the $1,2$-addition product.
The product formed is $3$-chloro-$1$-butene.

(C) The stability of alkenes is determined by the number of alkyl groups attached to the double-bonded carbons (hyperconjugation) and steric hindrance.
$1$-Butene is a monosubstituted alkene,making it the least stable.
Between cis-$2$-Butene and trans-$2$-Butene,both are disubstituted,but trans-$2$-Butene is more stable than cis-$2$-Butene due to lower steric repulsion between the methyl groups.
Therefore,the order of stability is: trans-$2$-Butene > cis-$2$-Butene > $1$-Butene.

(C) The debromination of vicinal dihalides using zinc dust $(Zn)$ in an alcoholic solvent follows an $E2$ elimination mechanism.
For meso-$2,3$-dibromobutane,the molecule can adopt a conformation where the two bromine atoms are anti-periplanar to each other.
Upon elimination of $ZnBr_2$,the anti-elimination process leads to the formation of trans-$2$-butene as the major product because the methyl groups are positioned opposite to each other in the transition state.

(D) Indirect hydration of an alkene involves the addition of water across the double bond,typically following Markovnikov's rule.
$Methyl \ alcohol$ $(CH_3OH)$ cannot be prepared by the indirect hydration of any alkene because the simplest alkene,$ethene$ $(CH_2=CH_2)$,produces $ethanol$ $(CH_3CH_2OH)$ upon hydration.
Since $methyl \ alcohol$ contains only one carbon atom,it is impossible to obtain it from an alkene,as the smallest alkene contains two carbon atoms.

(D) Markovnikov's rule is applicable to unsymmetrical alkenes.
In an unsymmetrical alkene,the negative part of the addendum $(Br^-)$ attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms.
Among the given options,$CH_3 - CH = CH_2$ (propene) is an unsymmetrical alkene,whereas $CH_2 = CH_2$ and $CH_3 - CH = CH - CH_3$ are symmetrical alkenes.
Therefore,the correct answer is $CH_3 - CH = CH_2$.

(B) The reaction of $ICl$ with propene follows Markovnikov's rule,where the electrophile $I^+$ attacks the double bond to form the more stable carbocation.
Since $I$ is less electronegative than $Cl$,$ICl$ polarizes as $I^{\delta+} - Cl^{\delta-}$.
The $I^+$ electrophile adds to the terminal carbon $(C_1)$ to form a more stable secondary carbocation at $C_2$.
Then,the nucleophile $Cl^-$ attacks the $C_2$ carbocation.
Thus,the major product is $2$-chloro-$1$-iodopropane.

(A) The conversion of $1$-butene $(CH_3CH_2CH=CH_2)$ to butane $(CH_3CH_2CH_2CH_3)$ is a hydrogenation reaction.
Hydrogenation of alkenes is typically carried out using hydrogen gas $(H_2)$ in the presence of a metal catalyst such as $Pd$,$Pt$,or $Ni$.
$H_2 / Pd$ is a standard reagent for the catalytic hydrogenation of alkenes to alkanes.
$Zn - HCl$ and $Sn - HCl$ are typically used for the reduction of nitro compounds to amines.
$Zn - Hg / HCl$ is the Clemmensen reduction reagent,used for the reduction of carbonyl groups $(C=O)$ to methylene groups $(CH_2)$.