Chemical Reactions Of Alcohols Phenols And Ethers

Hypothyroidism is a condition that occurs when the thyroid gland does not produce enough thyroid hormones, primarily thyroxine. Thyroxine is essential for regulating metabolism and other bodily functions. Deficiency of thyroxine leads to a slower metabolism and various symptoms associated with hypothyroidism.

The correct option is (D): Thyroxine

In the first step, bromine reacts with nitrobenzene in the presence of FeBr₃, forming a bromonitrobenzene derivative (product ).
In the second step, sodium nitrite ( ) at 273 K reacts with the product to form a diazonium salt ( ).
In the third step, the diazonium salt reacts with dilute HCl to produce a phenol derivative.
Finally, when the reaction mixture is warmed with water, the diazonium group is replaced by a hydroxyl group, forming as the final product, which corresponds to option (C).

Thus, the final product is .

This reaction is a type of Aldol condensation reaction. The reaction involves the condensation of benzaldehyde ( ) and acetaldehyde ( ) in the presence of dilute NaOH. The condensation leads to the formation of an enolate ion, which then reacts with another molecule of aldehyde.

The reaction produces a product in which the aldehyde group from one molecule reacts with the carbon-carbon bond from another molecule to form a conjugated product. The product of this reaction is cinnamaldehyde ( ), which has both an aldehyde group and a double bond.

Thus, the product 'P' is cinnamaldehyde.

The correct option is : (D)

The strength of a base is determined by the availability of the lone pair of electrons on the anion, which can accept a proton. Stronger bases have a higher tendency to donate electrons and accept protons.

- (methoxide ion) is a strong base because the negative charge is on a small, highly electronegative oxygen atom with no stabilizing groups attached.

- (acetate ion) is weaker than because the negative charge is delocalized between the two oxygen atoms, stabilizing the ion and reducing its basicity.

- (chloride ion) is a very weak base because the negative charge is highly stabilized by the electronegativity of chlorine, making it unlikely to accept a proton.

- (hydroxide ion) is a strong base, but it is weaker than because oxygen is less electron-rich in hydroxide compared to methoxide.

Thus, the strongest base among the given options is , which corresponds to option (A).

Let's evaluate each reaction to see if it results in the formation of an ether:

1. For P: C H CH Br + CH ONa:
- This is a nucleophilic substitution reaction. The CH O (methoxide ion) attacks the electrophilic carbon attached to the bromine (C H CH Br), resulting in the formation of an ether (C H CH OCH ). - Therefore, this reaction gives an ether as the major product.

2. For Q: C H ONa + CH Br:
- This is also a nucleophilic substitution. The phenoxide ion (C H O ) attacks the methyl group (CH ) attached to bromine, forming phenyl methyl ether (C H OCH ). - Therefore, this reaction also gives an ether as the major product.

3. For R: (CH ) C–Cl + CH ONa:
- In this case, the methoxide ion (CH O ) attacks the tertiary carbon of the alkyl chloride (C Cl). However, because this is a tertiary carbocation intermediate, the reaction is more likely to result in a rearrangement or a different product rather than an ether. - Therefore, this reaction does not give an ether as the major product.

4. For S: C H CH = CHCl + CH ONa:
- The CH O will attack the vinyl chloride (C H CH = CHCl), but this is not likely to form an ether under normal conditions due to the nature of the vinyl group.
Therefore, this reaction does not give an ether as the major product. Thus, the reactions P and Q both lead to the formation of an ether as the major product.

So, the correct answer is (A): Both P and Q.

The SN reaction involves a two-step mechanism:
1. The first step is the formation of a carbocation after the leaving group departs.
2. The second step is the attack by the nucleophile. For the SN reaction, the reactivity depends on the stability of the carbocation formed. A more stable carbocation leads to higher reactivity.
In option (A) , the phenyl group is attached to the halide, and the carbocation formed would be unstable due to the lack of resonance stabilization.
In option (B) , the phenyl group can provide some resonance stabilization to the carbocation formed, making it more stable and reactive towards the SN mechanism.
In options (C) and (D), the alkyl halides would form carbocations that are less stable compared to the phenylmethyl carbocation in option (B). Thus, shows the highest reactivity towards the SN reaction.

The correct option is (B) :

The complex involves a coordination compound with cobalt(III) at the center, where is ethylenediamine, a bidentate ligand, and is a monodentate ligand. Since ethylenediamine is bidentate, it can form chelate rings, and the position of the chloride ions relative to the chelate rings gives rise to stereoisomers.
The number of stereoisomers can be determined based on the possible geometric arrangements of the chloride ions and the ethylenediamine ligands. For a complex like , there are two possible isomers:
Cis-isomer : Both chloride ions are adjacent to each other.
Trans-isomer : Chloride ions are opposite each other. Additionally, since the two ethylenediamine ligands can occupy two positions (cis and trans), this leads to three stereoisomers in total.
Thus, the complex exhibits 3 stereoisomers.

The correct option is (A) : 3

1. For the complex ion [Fe(C O ) ] :
The oxalate ion (C O ) is a bidentate ligand, meaning it forms two bonds with the central metal ion.
Since there are 3 oxalate ions in the complex, each forming two bonds with Fe, the total coordination number of Fe is .
2. For the complex ion [Co(SCN) ] :
The thiocyanate ion (SCN ) is a monodentate ligand, meaning it forms only one bond with the central metal ion.
Since there are 4 SCN ions in the complex, the coordination number of Co is 4.
Thus, the coordination numbers are: Fe: 6, Co: 4

The correct option is (A): 6 and 4

Enzyme inhibitors, allosteric sites, and antagonists all play roles in drug-enzyme interactions. Enzyme inhibitors block the active site or allosteric site, preventing the enzyme from functioning. Antagonists interfere with the enzyme's normal activity. Co-enzymes, however, are organic molecules that assist enzymes in catalyzing reactions but are not directly involved in drug-enzyme interactions.

The correct option is (D): Co-enzymes

BHA (Butylated HydroxyAnisole) is a synthetic antioxidant commonly used as a food additive to prevent the oxidation of fats and oils in food products. Oxidation can lead to rancidity and spoilage, and antioxidants like BHA help preserve the quality and shelf life of foods.

The correct option is (B): BHA