Isomerism

Answer (d) (ii), (iii)

But-2-ene shows cis-trans isomerism. These are as following:

while other compounds pent-1-ene, but-1-ene, propene and ethene do not show cis-tr??s isomerism.

Answer (d) Ethylene glycol

We need to match the coordination compounds from Column I with their correct type of isomerism from Column II.
- (a) : This compound exhibits geometrical isomerism as it has two different ligands ( and ) arranged in different positions. Hence, it is related to (iv) geometrical isomerism.
- (b) : The compound with ethylenediamine ( ) as the ligand exhibits optical isomerism due to the possibility of non-superimposable mirror images. Hence, it matches with (iii) optical isomerism.
- (c) : This compound exhibits linkage isomerism because the sulfate ion ( ) can bind through either the sulfur or oxygen atom. Hence, it corresponds to (ii) linkage isomerism.
- (d) : This compound exhibits ionisation isomerism because it can form different ions when dissolved, depending on the position of the chloride and nitrate ions. Hence, it matches with (i) ionisation isomerism.
Thus, the correct matching is: (a) , (b) , (c) , (d) .
Thus, the correct answer is option (C), (a)-(iv), (b)-(iii), (c)-(ii), (d)-(ii).

Step 1: Optical isomerism occurs when a molecule has a chiral center, i.e., a carbon atom attached to four different substituents. For alkenes, this can occur if the double bond leads to non-superimposable mirror images (enantiomers).

Step 2: To check for optical isomerism, we examine each option:

- Option (A) : This molecule has no chiral centers because the two substituents on the second carbon (the methyl groups) are identical, preventing optical isomerism.

- Option (B) : This also lacks a chiral center due to the symmetry of the double bond, so it does not exhibit optical isomerism.

- Option (D) : This compound also does not have a chiral center because the substituents on the carbon atom with the double bond are not different enough to create optical isomerism.

- Option (E) : This is an alkane, not an alkene, and lacks a double bond, so it cannot exhibit optical isomerism.

Step 3: Option (C) has a chiral center at the third carbon due to the different substituents attached to it: a hydrogen atom, a methyl group, and two different alkyl groups. This makes the molecule optically active and capable of exhibiting optical isomerism.

Thus, the correct answer is option (C).

A molecule is considered chiral if it has a non-superimposable mirror image, which generally requires the presence of a chiral center. A chiral center is typically a carbon atom bonded to four different groups. If a molecule has a plane of symmetry or superimposable mirror images, it is achiral. - Option (A): 2-Chlorobutane is a chiral molecule. The second carbon is bonded to four different groups, making it a chiral center. - Option (B): 2,3-Dihydroxy propanal is a chiral molecule because both the second and third carbon atoms are each attached to four different groups, which makes the molecule chiral. - Option (C): 2-Bromo propionic acid is also chiral because the central carbon (C-2) has four different substituents, making it a chiral center. - Option (D): Butan-2-ol is a chiral molecule. The second carbon is attached to four different groups, so it is a chiral center. - Option (E): 2-Bromo-2-methoxypropane is not chiral. The central carbon (C-2) is attached to two identical groups (a bromine and a methoxy group), making it a symmetrical center. As a result, this molecule has a plane of symmetry and is achiral. Thus, the correct answer is (E) 2-Bromo-2-methoxypropane, which is not chiral.

Step 1: Understanding the Concept:
Optical activity is shown by chiral molecules, which typically have a carbon atom bonded to four different substituents (a chiral center or stereocenter).

Step 2: Detailed Explanation:
Examine each compound for a chiral carbon:
• 1-Chlorobutane (CH CH CH CH Cl): C1 is bonded to Cl, H, and two identical groups (CH CH CH and H? No, C1 has Cl, H, and a CH CH CH group. Wait, C1 is -CH Cl, so it has two H atoms. Not chiral.
• neo-Pentyl chloride ((CH ) C-CH Cl): The carbon with the Cl is a -CH Cl group, not chiral. The central quaternary carbon has four identical methyl groups? It has three methyls and one CH Cl. All four groups are not different. Wait, (CH ) C-CH Cl: The central carbon (C) is bonded to three CH groups and one CH Cl group. Three of the groups are identical (CH ), so it's not a chiral center. The CH Cl carbon has two H atoms.
• Isobutyl chloride ((CH ) CH-CH Cl): The CH Cl carbon has two H atoms. The CH carbon is bonded to two methyl groups, an H, and a CH Cl group. Two of the groups are identical (CH ), so not chiral.
• tert-Butyl chloride ((CH ) C-Cl): The central carbon is bonded to three identical methyl groups. Not chiral.
• 2-Chlorobutane (CH -CHCl-CH CH ): The carbon bearing the chlorine (C2) is bonded to four different groups: a chlorine atom, a hydrogen atom, a methyl group (CH -), and an ethyl group (-CH CH ). This is a chiral center, and the compound is optically active.

Step 3: Final Answer:
2-Chlorobutane is optically active.

Step 1: Understanding the Question:
The question asks to classify the given halogen-containing compound based on the position of the halogen atom (X).
Structure: Benzene-CH=CH-CH(X)-CH CH

Step 2: Detailed Explanation:
Let's analyze the position of the halogen atom X:
1. The halogen X is bonded to an hybridized carbon atom.
2. This carbon is directly adjacent to a carbon-carbon double bond ( ).
By definition:
- Vinylic halide: Halogen is bonded to an carbon of a bond.
- Allylic halide: Halogen is bonded to an carbon adjacent to a bond.
- Benzylic halide: Halogen is bonded to an carbon adjacent to an aromatic ring.
- Aryl halide: Halogen is bonded directly to the aromatic ring.
In the given structure, X is on a carbon adjacent to a group. This is an allylic position.

Step 3: Final Answer:
The compound is an allylic halide.

Concept:
Cross aldol condensation between an aromatic aldehyde and ketone forms -unsaturated ketone.
Step 1: Identify product structure.
Given product: This is chalcone.
Step 2: Recognize formation of chalcone.
Chalcone is prepared by Claisen-Schmidt condensation of:
Step 3: Reason.
- Benzaldehyde has no -hydrogen. - Acetophenone forms enolate ion. - Condensation followed by dehydration gives chalcone.
Step 4: Final answer.

Hence, correct option is (B).

Concept:
Ozonolysis breaks the double bond of an alkene and converts each double bonded carbon into carbonyl compounds.
Step 1: Write the products.
Products formed are: and
Step 2: Locate carbonyl carbons.
During ozonolysis, the carbonyl carbons were originally connected by a double bond in the alkene.
Step 3: Join both fragments.
Removing oxygen and joining carbonyl carbons: This corresponds to:
Step 4: Final answer.

Hence, correct option is (B).

Concept:
Structural isomers have same molecular formula but different connectivity of atoms.
Step 1: Parent carbon chain.
For , the parent chain is propane with two hydrogen atoms replaced by two chlorine atoms.
Step 2: Write all possible positions of two chlorine atoms.
Distinct arrangements are:
Step 3: Check duplicates.
By chain symmetry, no other new arrangement is possible.
Step 4: Total isomers.

Hence, correct option is (D).

Concept: Chemistry - Molecular Orbital Theory.
Step 1: Check bond type in each molecule. * has single bond. * has bond order 1. * has one and one bond. * is unstable. * has bond order 2.
Step 2: Bonding in . For , molecular orbital configuration shows two bonds and no net bond. Thus it behaves as a double bond made of: [ \pi + \pi ]
Step 3: Final answer. [ \boxed{C_2} ]
Hence, correct option is (D).

Concept: Chemistry - Heat Capacity. [ Q=nC\Delta T ]
Step 1: Find number of moles. [ n=\frac{60}{30}=2 ]
Step 2: Find temperature rise. [ \Delta T=54-30=24K ]
Step 3: Calculate heat required. [ Q=nC\Delta T ] [ Q=2 \times 25 \times 24=1200\text{ J} ]
Step 4: Convert into kJ. [ 1200\text{ J}=1.20\text{ kJ} ]
Step 5: Final answer. [ \boxed{1.20\text{ kJ}} ]
Hence, correct option is (B).

Step 1: Understanding the Concept:
Structural isomers are molecules with the same molecular formula but different bonding patterns (connectivity). To find them systematically, first determine the degree of unsaturation, draw the carbon skeleton, and then place the functional groups (chlorine atoms) in all possible unique positions.

Step 2: Key Formula or Approach:
Calculate the Degree of Unsaturation (DU) to confirm there are no rings or double bonds.

.
Since DU is 0, the carbon skeleton is an open-chain alkane.

Step 3: Detailed Explanation:
The basic carbon skeleton for a 3-carbon alkane is propane: .
We need to attach two chlorine atoms to this skeleton. We can classify the possibilities based on whether the chlorine atoms are on the same carbon (geminal) or different carbons (vicinal/terminal).
Case 1: Both Cl atoms on the same carbon (Geminal dihalides)
- Place both on an end carbon (C1):
Name: 1,1-dichloropropane
- Place both on the middle carbon (C2):
Name: 2,2-dichloropropane
Case 2: Cl atoms on different carbons
- Place them on adjacent carbons (C1 and C2):
Name: 1,2-dichloropropane
- Place them on the terminal carbons (C1 and C3):
Name: 1,3-dichloropropane
There are no other unique ways to attach the two chlorine atoms to a 3-carbon chain.

Step 4: Final Answer:
There are 4 possible structural isomers.