Class 11 Chemistry Worksheet on Chapter 12 Organic Chemistry - Some Basic Principles & Techniques - Set 3

Here are the solutions to the chemistry worksheet problems, broken down step-by-step.

Q1. How many pi bonds are there in phosgene?
* Reasoning: Phosgene has the chemical formula $COCl_2$. The central Carbon atom forms a double bond with Oxygen ($C=O$) and single bonds with two Chlorine atoms ($C-Cl$). A double bond consists of one sigma ($\sigma$) bond and one pi ($\pi$) bond. Single bonds are only sigma bonds. Therefore, there is only 1 pi bond in the molecule (in the C=O bond).
* Answer: (a) One

Q2. What is the IUPAC name of $CH_3OC_2H_5$?
* Reasoning: This is an ether. In IUPAC nomenclature for ethers, we take the smaller alkyl group attached to the oxygen as a substituent ending in "-oxy" and the larger group as the parent chain. Here, Methyl ($CH_3$) is smaller than Ethyl ($C_2H_5$). So, "Methoxy" is the prefix and "ethane" is the suffix. The name is Methoxy ethane.
* Answer: (b) Methoxy ethane

Q3. What is the colour of the bromine solution?
* Reasoning: Bromine water or a solution of bromine in organic solvents like carbon tetrachloride typically appears reddish-brown. Among the given options, Red is the correct description.
* Answer: (c) Red

Q4. Which of the following functional groups is present in RCOOH?
* Reasoning: The general formula $RCOOH$ represents a Carboxylic acid. It contains the carboxyl group ($-COOH$), which consists of a carbonyl group ($C=O$) and a hydroxyl group ($-OH$) attached to the same carbon.
* Answer: (b) Carboxylic acid

Q5. What is the suffix of esters?
* Reasoning: Esters are named as alkyl alkanoates. The part derived from the acid ends with the suffix -oate. For example, methyl ethanoate.
* Answer: (a) oate

Q6. Draw the bond-line formula of heptan-4-one.
* Reasoning:
* "Hept" means a 7-carbon chain.
* "an" means all single bonds between carbons.
* "4-one" means there is a ketone group ($C=O$) on the 4th carbon.
* Structure: A zigzag line with 7 points (carbons). The double bond to Oxygen is on the middle point (carbon #4).

Q7. What are the hybridisation states of each carbon in $CH_3 - CH = CH - CN$?
* Reasoning: Let's look at each carbon from left to right:
1. $CH_3-$: This carbon has 4 single bonds (3 to H, 1 to C). Steric number 4 $\rightarrow$ $sp^3$.
2. $-CH=$: This carbon has a double bond. Steric number 3 (1 single to C, 1 double to C, 1 to H) $\rightarrow$ $sp^2$.
3. $=CH-$: This carbon also has a double bond. Steric number 3 $\rightarrow$ $sp^2$.
4. $-CN$: This is a nitrile group. The carbon has a triple bond to Nitrogen and a single bond to Carbon. Steric number 2 $\rightarrow$ $sp$.
* Answer: $sp^3, sp^2, sp^2, sp$

Q8. Which of the following carbocation is most stable?
* Reasoning: Carbocation stability increases with the number of alkyl groups attached to the positively charged carbon (due to the inductive effect and hyperconjugation). The order is Tertiary ($3^\circ$) > Secondary ($2^\circ$) > Primary ($1^\circ$) > Methyl.
* $(CH_3)_3CCH_2^+$: Primary ($1^\circ$)
* $(CH_3)_3C^+$: Tertiary ($3^\circ$) — The positive charge is on a carbon attached to 3 other methyl groups.
* $CH_3CH_2CH_2^+$: Primary ($1^\circ$)
* $CH_3CH^+CH_3$: Secondary ($2^\circ$)
* The tertiary carbocation is the most stable.
* Answer: $(CH_3)_3C^+$

Q9. Draw the resonating structure of phenol and nitrobenzene.
* Reasoning:
* Phenol: The lone pair on the Oxygen atom delocalizes into the benzene ring. This creates negative charges at the ortho (2, 6) and para (4) positions. You draw arrows moving electrons from O into the ring, shifting the double bonds around the ring.
* Nitrobenzene: The nitro group ($-NO_2$) withdraws electrons. The pi electrons from the benzene ring move towards the nitro group. This creates positive charges at the ortho and para positions.

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Final Answer:

Q1: (a) One
Q2: (b) Methoxy ethane
Q3: (c) Red
Q4: (b) Carboxylic acid
Q5: (a) oate
Q6: [See reasoning above for drawing instructions: 7-carbon chain with =O on the 4th carbon]
Q7: $sp^3, sp^2, sp^2, sp$
Q8: $(CH_3)_3C^+$
Q9: [See reasoning above: Phenol shows electron donation to ortho/para positions; Nitrobenzene shows electron withdrawal creating positive charges at ortho/para positions]