Answer:
(1)NBS, ROOR (2)
and (3)HBr, ROOR
Explanation:
Conversion of ethylbenzene to (2-bromoethyl)benzene involves three steps.
First step: allylic bromination of ethylbenzene in presence of NBS, ROOR. This step produces (1-bromoethyl)benzene.
Second step: E2 ellimination of HBr from (1-bromoethyl)benzene in presence of a strong base e.g. potassium t-butoxide []. This step produces styrene
Third step: HBr addition to double bond in presence of HBr, ROOR (antimarkonikov addition of HBr). Thus step produces (2-bromoethyl)benzene.
Full reaction scheme has been shown below.
D is asking basically asking if would you get the same results if the tubes had salt instead of sugar
Both transition metals and alkali metals are good conductors of heat and electricity, react with water, and are easily oxidized.
<h3>What are alkali metals and transition metals?</h3>
The alkali metals are elements of group 1 which are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr). They are also known as the s-block elements because they have their outermost electron in an s-orbital.
The alkali metals are shiny, soft, highly reactive metals and readily lose their outermost electron to create cations with charge +1. They can tarnish rapidly in the air due to oxidation by atmospheric moisture and oxygen.
Transition elements or transition metals are elements that have partially filled d-orbitals. An element having a d-subshell that is partially filled with electrons or can form stable cations with an incompletely filled d orbital.
Any element present in the d-block of the modern periodic table which consists of groups 3 to 12, is considered to be a transition element. For example, the mercury in the +2 oxidation state, corresponds to an electronic configuration of (n-1)d¹⁰. Many paramagnetic compounds are formed by transition metals because they have unpaired electrons in the d orbital.
Learn more about transition metals and alkali metals, here:
brainly.com/question/15775417
#SPJ1
