Give the IUPAC name for each compound.

Chemistry

1 answer:

Kaylis [27]3 years ago

3 0

Answer:

a. 1-fluoro-3,3,4-trimethyl-pentane.

b. 1-iodo-3-ethyl-2-methyl-hexane.

c. 1,3-dichloro-5-dimethyl-hexane.

d. 1-bromo-3-methyl-cyclopentane.

Explanation:

Hello!

In this case, according to the IUPAC rules for the listed alkyl halides, we first need to name the halogens (considering periodic order) then alkyl radicals and finally the parent chain; thus, the names are given below:

a. 1-fluoro-3,3,4-trimethyl-pentane.

b. 1-iodo-3-ethyl-2-methyl-hexane.

c. 1,3-dichloro-5-dimethyl-hexane.

d. 1-bromo-3-methyl-cyclopentane.

e. 5-chloro-1-bromo-1,1,5-trimethyl-pentane (radicals are not clear).

Best regards!

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For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

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Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

$\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)$ .

When $\rm H_2$ is added to $\rm H_2C\text{=}CH_2$ (ethene) under heat and with the presence of a catalyst, $\rm H_3C\text{-}CH3$ (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, $\Delta S < 0$ .

The equation for the change in Gibbs Free Energy for a particular reaction is:

$\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})$ .

For a particular reaction, the more negative $\Delta G$ is, the more spontaneous ("favorable") the reaction would be.

Since typically $\Delta S < 0$ for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

$T$ (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of