GUJCET 2013 Chemistry Question Paper with Answer and Solution

(D) The given structure is $CH_3-CH=CH-CH_2Cl$.
$1$. Identify the longest carbon chain containing the double bond: The chain has $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain to give the lowest possible number to the double bond: Numbering from the right gives the double bond position at $C-2$.
$3$. Identify the substituent: There is a chloro group at $C-1$.
$4$. Combine these to get the $IUPAC$ name: $1-$chloro but$-2-$ene.

(A) The correct answer is $A$. Ethers are generally resistant to oxidation and reduction under normal conditions. They do not react with oxidizing agents like $KMnO_4$ or reducing agents like $LiAlH_4$ easily. Option $B$ is correct as ethers lack the ability to form intermolecular $H$-bonds with water,making them less soluble. Option $C$ is correct as it describes the dehydration of alcohols to form ethers. Option $D$ is correct as ethers cannot form intermolecular $H$-bonds with themselves,leading to lower boiling points compared to isomeric alcohols.

(D) The acidity of alcohols depends on the stability of the alkoxide ion formed after the loss of a proton $(H^+)$.
Alkyl groups are electron-donating groups ($+I$ effect).
As the number of alkyl groups attached to the carbon atom bearing the $-OH$ group increases,the electron density on the oxygen atom increases due to the $+I$ effect,which destabilizes the alkoxide ion.
Therefore,the order of acidity is: $primary > secondary > tertiary$.
Among the given options:
$(A)$ $(CH_3)_3C-OH$ is a tertiary alcohol.
$(B)$ $CH_3-CH_2-CH_2-OH$ is a primary alcohol.
$(C)$ $CH_3-CH(OH)-CH_3$ is a secondary alcohol.
$(D)$ $CH_3-OH$ is methanol (the simplest alcohol with no alkyl group attached to the carbon bearing $-OH$).
Since methanol has no electron-donating alkyl groups to destabilize the methoxide ion,it is the most acidic among the given options.

(A) The $Wolff-Kishner$ reduction is a chemical reaction used to convert carbonyl groups (aldehydes or ketones) into methylene groups $(-CH_2-)$ using hydrazine $(NH_2NH_2)$ and a strong base like $KOH$ in a high-boiling solvent.
$Acetone$ $(CH_3COCH_3)$ has $3$ carbon atoms. Upon $Wolff-Kishner$ reduction,it is converted into $Propane$ $(CH_3CH_2CH_3)$,which also contains $3$ carbon atoms.
Therefore,the correct option is $A$.

(B) Benzaldehyde is widely used in the perfume industry due to its characteristic almond-like odor.
It is an aromatic aldehyde used as a flavoring agent and in the synthesis of various organic compounds.
Therefore,the correct option is $B$.

(D) The acidity of substituted benzoic acids is influenced by the inductive effect $(-I)$ and resonance effect $(-R)$ of the substituents.
$2-$Nitrobenzoic acid is the strongest acid among the given options due to the ortho effect.
The ortho effect is a steric and electronic phenomenon where the presence of a substituent at the ortho position relative to the carboxyl group increases the acidity,regardless of whether the substituent is electron-withdrawing or electron-donating.
Therefore,the correct order of acidity is: $2-$Nitrobenzoic acid $> 3-$Nitrobenzoic acid $> 4-$Nitrobenzoic acid $>$ Benzoic acid.

(A) Nitrogenous bases are classified into two types: purines and pyrimidines.
Purines are bicyclic structures (a six-membered ring fused to a five-membered ring).
Pyrimidines are monocyclic structures (a single six-membered ring).
Among the given options,$Guanine$ and $Adenine$ are purines (bicyclic).
$Thymine$,$Uracil$,and $Cytosine$ are pyrimidines (monocyclic).
Therefore,$Guanine$ is the correct answer.

(B) The chemical name of vitamin $B_6$ is $Pyridoxine$.
Vitamin $B_6$ is a water-soluble vitamin that plays an important role in amino acid metabolism.
Therefore,the correct option is $B$.

(D) The reaction of glucose with $HI$ upon prolonged heating leads to the formation of $n-\text{hexane}$.
This reaction confirms that all $6$ carbon atoms in glucose are linked in a straight chain.
Therefore,the reaction $C_{6}H_{12}O_{6} \xrightarrow{HI, \Delta} n-\text{hexane}$ indicates the linear structure of glucose.
Thus,the correct option is $D$.

(A) The rate law is given by $v_1 = K[A]^2[B]^0 = K[A]^2$.
When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is doubled $([B]' = 2[B])$,the new rate $v_2$ is:
$v_2 = K(2[A])^2(2[B])^0$
$v_2 = K(4[A]^2)(1)$
$v_2 = 4 \times K[A]^2$
$v_2 = 4 \times v_1$.
Therefore,the rate of reaction becomes $4$ times the initial rate.