I can't figure out the answer please help.
1. Which of the following would you expect to have the weakest C-X bond?
A)CH3Cl
B)CH3CH2Br
C)CH3F
D)CH3CH2I
E)(CH3)2CHBr
2. Which of the following haloalkanes would not undergo the reaction below?
R-X + CH3S^ ----> CH3SR + X^-
A)(CH3)2CHI
B)CH3Cl
C)CH3CH=CHBr
D)CH3CH2Br
E)CH3CH2CH2I
Here are a couple of clues (not answers):
1. One of the choices involves a Br atom bonded to a secondary carbon atom (one attached to two other carbon atoms). This is the one that forms the more stable carbonium ion, and the weakest C-Br bond.
2. CH3S^- would react most readily with the haloalkane that forms the least stable carbonium (R+) ion. Inversely, a primary haloalkane with the shortest carbon chain would be the least likely to react.
loihjuiu
To determine the answer to question 1, which asks about the weakest C-X bond, you need to consider the electronegativity difference between carbon (C) and the halogen (X) in each compound. Generally, the weaker the C-X bond, the greater the electronegativity difference between C and X.
Here's a step-by-step approach to figuring out the answer:
1. Look at each compound and identify the halogen atom (X) bonded to the carbon (C) atom.
2. Look up the electronegativity values of the halogens and carbon.
- Carbon (C) has an electronegativity value of approximately 2.5.
- The halogens' electronegativity values are as follows:
- Fluorine (F): approximately 3.98
- Chlorine (Cl): approximately 3.16
- Bromine (Br): approximately 2.96
- Iodine (I): approximately 2.66
3. Calculate the electronegativity difference for each compound by subtracting the electronegativity of carbon from the electronegativity of the halogen.
Now, let's go through the options:
A) CH3Cl: The electronegativity difference is approximately 3.16 - 2.5 = 0.66.
B) CH3CH2Br: The electronegativity difference is approximately 2.96 - 2.5 = 0.46.
C) CH3F: The electronegativity difference is approximately 3.98 - 2.5 = 1.48.
D) CH3CH2I: The electronegativity difference is approximately 2.66 - 2.5 = 0.16.
E) (CH3)2CHBr: The electronegativity difference is approximately 2.96 - 2.5 = 0.46.
Based on the electronegativity differences calculated above, the compound with the weakest C-X bond would be the one with the lowest electronegativity difference. In this case, it is option D) CH3CH2I with an electronegativity difference of approximately 0.16.
To answer the second question, we need to determine which haloalkane would not undergo the reaction given:
R-X + CH3S^ ----> CH3SR + X^-
In this reaction, the R-X undergoes a substitution reaction with CH3S^ to form CH3SR and X^-.
To figure this out, we need to consider the leaving group ability of X. The better the leaving group, the more likely the substitution reaction will occur.
Here's a step-by-step approach:
1. Look at each compound and identify the halogen atom (X).
2. Evaluate the halogen's ability to function as a leaving group based on certain factors, such as its size and electronegativity.
- Smaller halogens are generally better leaving groups.
- Halogens with higher electronegativity tend to be poorer leaving groups.
3. Identify the compound where the halogen serves as a poorer leaving group, suggesting that the reaction may not occur.
Let's assess the options:
A) (CH3)2CHI: Iodine (I) is a good leaving group due to its size and lower electronegativity.
B) CH3Cl: Chlorine (Cl) is a good leaving group due to its smaller size.
C) CH3CH=CHBr: Bromine (Br) is a good leaving group as it's not highly electronegative.
D) CH3CH2Br: Bromine (Br) is a good leaving group due to its smaller size.
E) CH3CH2CH2I: Iodine (I) is a good leaving group.
Based on the analysis, the compound that would not undergo the given reaction is option C) CH3CH=CHBr since bromine is a good leaving group and the reaction is likely to occur in this case.
Remember, understanding the concepts behind these questions will help you tackle similar problems in the future!
Sorry, my clue to the second question needs correction:
2. CH3S^- would react most readily with the haloalkane that forms the MOST stable carbonium (R+) ion. Inversely, a primary haloalkane with the shortest carbon chain would be the least likely to react.