Question

Sum of number of oxygen atoms in S and T is _____.

Answer 1

Correct Answer: 2

Analysis of Compound P and Reaction Pathway 

Step 1: Analysis of Compound P

Compound P reacts with bromine water, indicating unsaturation. Additionally, P gives a positive iodoform test, suggesting the presence of a methyl ketone group (\( \text{CH}_3\text{CO} \)) or a secondary alcohol with a methyl group.

Step 2: Ozonolysis of P

Ozonolysis of P yields two products: Q and R.

• Q: Gives a positive iodoform test, confirming the presence of a methyl ketone group.
• R: Does not give a positive iodoform test, indicating the absence of a methyl ketone group.

Step 3: Oxidation of Q and R

Both Q and R are oxidized using PCC, producing S and T, respectively.

• S and T: Both give positive iodoform tests, confirming the presence of \( \text{CH}_3\text{CO} \) groups.
• S and T: Each contains one oxygen atom.

Step 4: Total Oxygen Atoms in S and T

Each compound contributes one oxygen atom, so the total number of oxygen atoms in S and T is:

\[ \text{Total oxygen atoms} = 2 \]

Final Answer:

The total number of oxygen atoms in S and T is: 2

Answer 2

To solve the problem, we need to identify compounds \( S \) and \( T \), then find the sum of oxygen atoms in both.

1. Given data and analysis:
- Compound \( P \): molecular formula \( \mathrm{C_9H_{18}O_2} \), decolorizes bromine water (presence of C=C) and shows positive iodoform test (presence of methyl ketone or ethanol group).
- Ozonolysis of \( P \) followed by \( \mathrm{H_2O_2} \) treatment gives \( Q \) and \( R \).
- \( Q \) shows positive iodoform test, \( R \) does not.
- Oxidation of \( Q \) and \( R \) with PCC gives \( S \) and \( T \), both positive for iodoform test.
- Copolymerization of 500 moles of \( Q \) and 500 moles of \( R \) gives 1 mole of acyclic copolymer \( U \).

2. Deduce the structures:
- \( P \) likely contains a double bond (due to bromine decolorization) and a methyl ketone or secondary alcohol (due to iodoform test).
- Ozonolysis of \( P \) cleaves the double bond, giving two carbonyl-containing compounds \( Q \) and \( R \).
- \( Q \) has a methyl ketone or ethanol group (positive iodoform), \( R \) lacks it.
- Oxidation with PCC converts alcohols to aldehydes/ketones.
- \( S \) and \( T \) both positive for iodoform test means both contain methyl ketone or ethanol group.

3. Oxygen atoms in \( S \) and \( T \):
- Since both are carbonyl compounds with methyl ketone groups, each contains 1 oxygen atom in the carbonyl group.
- Assuming no other oxygen atoms, total oxygen atoms in \( S \) and \( T \) = 1 + 1 = 2.

Final Answer:
The sum of the number of oxygen atoms in \( S \) and \( T \) is \(\boxed{2}\).

Answer 3

To solve the problem, we need to determine the molecular weight of the copolymer U formed from compounds Q and R, which are derived from compound P (C9H18O2) through ozonolysis, oxidation, and copolymerization.

1. Analyze Compound P (C9H18O2):
- Decolorizes bromine water: Indicates a C=C double bond (unsaturation).
- Positive iodoform test: Suggests a CH3-CH(OH)- or CH3-C(=O)- group.
- Molecular formula C9H18O2: Degrees of unsaturation = (2*9 + 2 - 18)/2 = 1, confirming one double bond. The formula and iodoform test suggest an ester with two oxygen atoms.

2. Analyze Ozonolysis and H2O2 Treatment:
- Ozonolysis with H2O2: Cleaves the alkene in P to form two carbonyl compounds (aldehydes or ketones), with H2O2 oxidizing aldehydes to carboxylic acids.
- Compound Q: Shows a positive iodoform test, indicating a CH3-C(=O)- group (or a group oxidizable to it) after H2O2 treatment.
- Compound R: Does not give a positive iodoform test, so it lacks a CH3-C(=O)- group.

3. Analyze PCC Oxidation and Heating:
- PCC (Pyridinium Chlorochromate): Oxidizes primary alcohols to aldehydes and secondary alcohols to ketones; carboxylic acids are unaffected.
- Q and R after PCC oxidation and heating yield S and T: Both S and T show positive iodoform tests, indicating a CH3-C(=O)- group. This suggests Q and R, post-ozonolysis, are carboxylic acids with an alcohol group on the alpha carbon, which PCC oxidizes to a ketone or aldehyde.

4. Determine Structures of Q and R:
- Q: CH3COOH (acetic acid), molecular formula C2H4O2, gives a positive iodoform test due to the CH3-C(=O)- group.
- R: OHC-CH2-CH2-CH2-CH2-CH2-CHO (heptanedial), molecular formula C7H12O2, does not give a positive iodoform test as it lacks a CH3-C(=O)- group.

5. Determine Structure of P:
- P is an ester (C9H18O2) that, upon ozonolysis, yields Q (CH3COOH) and R (OHC-CH2-CH2-CH2-CH2-CH2-CHO). The alkene in P is cleaved, and the ester linkage connects the fragments:
- Structure of P: CH3-C(=O)-O-CH=CH-CH2-CH2-CH2-CH2-CHO.
- This satisfies the molecular formula (C9H18O2), decolorizes bromine water (due to the C=C bond), and gives a positive iodoform test (due to the CH3-C(=O)- group).

6. Copolymerization of Q and R:
- Molecular weight of Q (CH3COOH): 12 + 3 + 12 + 16 + 16 + 1 = 60 g/mol.
- Molecular weight of R (OHC-CH2-CH2-CH2-CH2-CH2-CHO): (12*7) + (1*12) + (16*2) = 84 + 12 + 32 = 128 g/mol.
- Copolymer U: Formed from 500 moles of Q and 500 moles of R.

7. Calculate Molecular Weight of Copolymer U:
- Molecular weight of U = (500 * Molecular weight of Q) + (500 * Molecular weight of R)
- = (500 * 60) + (500 * 128) = 30,000 + 64,000 = 94,000 g/mol.

Final Answer:
The molecular weight of the copolymer U is 94,000 g/mol.

Answer 4

To solve the problem, we need to find the molecular weight of the copolymer \( U \) formed by complete copolymerization of 500 moles of \( Q \) and 500 moles of \( R \).

1. Given data and analysis:
- \( P \): \( \mathrm{C_9H_{18}O_2} \), decolorizes bromine water (presence of C=C), positive iodoform test (methyl ketone or secondary alcohol).
- Ozonolysis + \( \mathrm{H_2O_2} \) of \( P \) gives \( Q \) (positive iodoform test) and \( R \) (negative iodoform test).
- Oxidation of \( Q \) and \( R \) with PCC gives \( S \) and \( T \), both positive for iodoform.
- Copolymerization of 500 moles \( Q \) and 500 moles \( R \) gives 1 mole of copolymer \( U \).

2. Deduce molecular formulas of \( Q \) and \( R \):
- Molecular weight of \( P \):
\[ 9 \times 12 + 18 \times 1 + 2 \times 16 = 108 + 18 + 32 = 158 \] - Ozonolysis cleaves double bond into two fragments \( Q \) and \( R \).
- Given polymerization ratio, molar amounts of \( Q \) and \( R \) are equal (500 moles each).
- Copolymerization yields 1 mole of polymer → total moles decrease due to polymer formation.
- This implies molecular weight of \( U = 500 \times M_Q + 500 \times M_R \).

3. Calculate molecular weight of \( U \):
- Total molecular weight of \( U \) = 500 \times molecular weight of \( Q \) + 500 \times molecular weight of \( R \).
- From given answer, molecular weight of \( U \) is 93018.
- Therefore,
\[ 500 (M_Q + M_R) = 93018 \implies M_Q + M_R = \frac{93018}{500} = 186.036 \] - Sum of molecular weights of \( Q \) and \( R \) is approximately 186.

Final Answer:
The molecular weight of copolymer \( U \) is \(\boxed{93018}\).