Answer:
The reagents are 
.
Explanation:
1-Methylenecyclopentene is treated with HBr form 1-bromo-1-methylcyclopentane, which is treated with strong base ethoxide ion and forms 1-methylcyclopent-1-ene.
This alkene is treated with osmium tetraoxide in the presence of sodium bisulfite to form target product.
The chemical reaction is as follows.
We have that the name (not chemical symbol) of the main group element in period 5 and group 3A is
Metalloid boron (B)
From the Question we are told that
The element belongs to
Period 5 and group 3A
Generally
Group 3A of the periodic table includes the metalloid boron (B), aluminum (Al), indium (In), and thallium (Tl) gallium (Ga),
Period 5 is possessed by the metalloid boron (B) of the Group 3A
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Answer:
The charge on H is partial positive and the charge on I is partial negative.
δ+H−Iδ−
The percent ionic character is 5.15 %.
Explanation:
The ionic compound is formed by the complete transfer of electron from one atom to another. The one atom thus carry the positive charge while the other which accept the electron became negative. like in case of NaCl .This have 100% ionic character.
In covalent bond, the bond is formed by the sharing of electron between two atoms. The one atom which greatly pull the electrons towards itself is became partial negative while the other became partial positive atom. In this way electronegativity difference is created. For example in HI bond hydrogen is partial positive and have 2.2 electronegativity while the electronegativity of iodine is 2.66.
Percent ionic character:
Formula:
%ionic character = [1- e∧-(0.25)(x)²] × 100
X= XA - XB
XA = higher electronegativity
XB = lower electronegativity
X = 2.66 - 2.2
X = 0.46
now we will determine the %ionic character.
%ionic character = [1- e∧-(0.25)(x)²] × 100
%ionic character = [1- e∧-(0.25)(0.46)²] × 100
%ionic character = [1- e∧-(0.25)(0.2116)] × 100
%ionic character = [1- e∧-(0.0529)] × 100
%ionic character = [1- 0.9485)] × 100
%ionic character = 0.0515 × 100
%ionic character = 5.15%
I have provided a pair of spectra, NMR and IR for one of the compounds in the question.
The NMR spectrum provides the least information, as there is only a single peak that is a singlet. All three compounds have a methyl group adjacent to a carbonyl that would appear as a singlet. However, the chemical shift is closer to 3 ppm which suggests it is being deshielded further compared to a standard carbonyl which would appear at 2 ppm. This suggests the structure is most likely 1,1,1-trichloropropanone.
Looking at the IR spectrum, we see the carbonyl stretch at approximately 1750 cm⁻¹. We also see the presence of strong stretches at 750 and 850 cm⁻¹ which are very characteristic of a C-Cl stretch. Therefore, there is enough evidence to suggest that these spectra correspond to 1,1,1-trichloropropanone.
The name of the compound is Thorium (IV) Iodide and formula of the compound ThI₄.
<h3>What is Molecular Formula ?</h3>
The chemical formula that gives total number of atoms of each element in one molecule of a compound is called Molecular Formula.
<h3>What is Atomic Number ?</h3>
The number of proton which is used to separate one element from the other element is called Atomic Number.
here the atomic of the element Q is 90 and the atomic number of R is 53. The element Q is Thorium. Symbol of Thorium is Th. The element R is Iodine. Symbol of Iodine is I.
The valency of thorium is +4 and valency of Iodine is -1. So the formula of the compound is ThI₄.
Thus from the above conclusion we can say that The name of the compound is Thorium (IV) Iodide and formula of the compound ThI₄.
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