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(a)
Interpretation:
The structure of alcohol that is required to prepare the given aldehyde has to be drawn.
Concept Introduction:
Carbonyl compounds such as
(b)
Interpretation:
The structure of alcohol that is required to prepare the given ketone has to be drawn.
Concept Introduction:
Carbonyl compounds such as aldehydes and ketones can be synthesized by oxidation or reduction reaction. When a primary alcohol undergoes oxidation reaction, the product obtained is carboxylic acid that is formed through the intermediate aldehyde. When mild oxidizing agent is used, aldehyde can be obtained as product from primary alcohol. When a secondary alcohol undergoes oxidation reaction, the product obtained is a ketone. This cannot be further oxidized. Tertiary alcohols do not undergo oxidation reaction. Some of the mild oxidizing agents used are
(c)
Interpretation:
The structure of alcohol that is required to prepare the given ketone has to be drawn.
Concept Introduction:
Carbonyl compounds such as aldehydes and ketones can be synthesized by oxidation or reduction reaction. When a primary alcohol undergoes oxidation reaction, the product obtained is carboxylic acid that is formed through the intermediate aldehyde. When mild oxidizing agent is used, aldehyde can be obtained as product from primary alcohol. When a secondary alcohol undergoes oxidation reaction, the product obtained is a ketone. This cannot be further oxidized. Tertiary alcohols do not undergo oxidation reaction. Some of the mild oxidizing agents used are
(d)
Interpretation:
The structure of alcohol that is required to prepare the given ketone has to be drawn.
Concept Introduction:
Carbonyl compounds such as aldehydes and ketones can be synthesized by oxidation or reduction reaction. When a primary alcohol undergoes oxidation reaction, the product obtained is carboxylic acid that is formed through the intermediate aldehyde. When mild oxidizing agent is used, aldehyde can be obtained as product from primary alcohol. When a secondary alcohol undergoes oxidation reaction, the product obtained is a ketone. This cannot be further oxidized. Tertiary alcohols do not undergo oxidation reaction. Some of the mild oxidizing agents used are
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Chapter 4 Solutions
Organic And Biological Chemistry
- true or falsegiven these two equilibria with their equilibrium constants:H2(g) + CI2(l) ↔ 2HCI(g) K= 0.006 CI2(l) ↔ CI2(g) K= 0.30The equilibrium contstant for the following reaction is 1.8H2(g) + CI2 ↔ 2HCI(g)arrow_forwardI2(g) + CI2(g) ↔ 2ICIK for this reaction is 81.9. Find the equilibrium concentration of I2 if the inital concentration of I2 and CI2 are 0.010 Marrow_forwardtrue or false,the equilibrium constant for this reaction is 0.50.PCI5(g) ↔ PCI3(g) + CI2(g)Based on the above, the equilibrium constant for the following reaction is 0.25.2PCI5(g) ↔. 2PCI3(g) + 2CI2(g)arrow_forward
- true or false, using the following equilibrium, if carbon dioxide is added the equilibrium will shift toward the productsC(s) + CO2(g) ↔ 2CO(g)arrow_forward2S2O2/3- (aq) + I2 (aq) ---> S4O2/6- (aq) +2I- (aq) Experiment I2 (M) S2O3- (M) Initital Rate (M/s) 1 0.01 0.01 0.0004 2 0.01 0.02 0.0004 3 0.02 0.01 0.0008 Calculate the overall order for this reaction using the table data a) 3b) 0c) 2d) 1arrow_forwardthe decomposition of N2O5 is the first order with a half-life of 1.98 minutes. If the inital concentration of N2O5 is 0.200 M, what is the concentration after 6 minutes?a) 0.612 Mb) 0.035 Mc) 0.024 Md) 0.100 Marrow_forward
- 20.00 mL of 0.150 M HCI is titrated with 0.075 M NaOH. What volume of NaOH is needed?a) 50 mLb) 20 mLc) 40 mLd) 26.66 mLarrow_forward20.00 mL of 0.150 M NaOH is titrated with 37.75 mL of HCI. What is the molarity of the HCI?a) 0.150 Mb) 0.079 Mc) 0.025 Md) 0.050 Marrow_forwardin the following reaction, the OH- acts as which of these?NO2- (aq) + H2O (l) ⇌ OH- (aq) + HNO2 (aq)a) not a weak acidb) basec) acidarrow_forward
- find the pH of a buffer made from 0.20 M HNO2 and 0.10 M NaNO2. Ka= 4.0 x 10-4a) 4.00b) 3.40c) 3.70d) 3.10arrow_forwardthe Ka for sodium dihydrogen phosphate is 6.32 x 10-8. Find the pH of a buffer made from 0.15 M H2PO4- and 0.15 M HPO42-.a) 6.98b) 7.42c) 7.00d) 7.20arrow_forwardFind the equilibrium concentration of H3O+ starting with 0.072 M solution of acetic acid. Ka = 1.8 x 10-5. Acetic acid is HC2H3O2 (aq).HC2H3O2 (aq) + H2O (l) ⇌ H3O (aq) + C2H3O2- (aq) a) 1.3 x 10-6 b) 1.1 x 10-3 c) 1.5 x 10-2 d) 3.6 x 10-5arrow_forward
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